New Simplified Chemistry Class 10 ICSE Solutions – Organic Chemistry

New Simplified Chemistry Class 10 ICSE Solutions – Organic Chemistry

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QUESTIONS
2004

 Question 1.
Write the equation for the preparation of ethylene from ethyl alcohol.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 1
Question 2.
State the general formula for a saturated hydrocarbon and give one example and structural formula of the same.
Answer:
General formula for a saturated hydrocarbon (alkanes): CnH2n+2 Example of saturated hydrocarbons (alkanes): C2H2×1+2 Or C1H4
Structural formula of CH4 (Methane)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 2

Question 3.
Name a compound which will give acetylene gas when treated with water.
Answer:
CaC2; (calcium carbide)

2005

Question 1.
Define the term ‘catenation’.
Answer:
It is the property of elements by virtue of which atoms of the element can link to each other to form chains or rings of different sizes.

Question 2.
State the term for: – Compounds having the same general formula, and similar chemical properties.
Answer:
Homologous series.

Question 3.
Draw the structural formula of a compound with two carbon atoms in each of the following cases:

  1. An alkane with a carbon to carbon single
  2. An alcohol containing two carbon atoms.
  3. An unsaturated hydrocarbon with a carbon to carbon triple bond

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 3

Question 4.
Ethane, Ethene, Ethanoic acid, Ethyne, Ethanol — From the compounds, name :

  1. The compound with — OH and with — COOH,
  2. Homologue of homologous series with general formula Cn H2n.

Answer:

  1. Compound with — OH group: Ethanol (C2H5OH)
  2. Compound with — COOH group: Ethanoic acid (CH3COOH)
  3. Homologue of homologous series with G.F. CnH2n:Ethene (C2H4)

Question 5.
Write the equations for the following laboratory preparations:

  1. Ethane from Sodium propionate.
  2. Ethene from Iodoethane.
  3. Ethyne from Calcium carbide.
  4. Methanol from Iodomethane.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 4

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 5

(3) CaC2 + 2H2O →HC≡ CH + Ca(OH)2

2006

Question 1.
Which one of the elements — Li, Be, B, C, O, F, Ne shows the property of catenation.
Answer:
C (carbon).

Question 2.
Write a balanced equation for:

  1. Reaction of ethane and oxygen in presence of molybdenum oxide.
  2. Preparation of CH4 from anhydrous sodium ethanoate (sodium acetate).
  3. Reaction of heating ethanol at 170° C in the presence of cone. H2SO4.
  4. Preparation of carbon tetrachloride from methane.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 6

Question 3.
Give the IUPAC name and the functional group for :

  1. CH3 — CH2 — CHO
  2. H3C — CH2 — CH2 — OH

Answer:
(1) IUPAC name: Propanal
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 7
(2) IUPAC name: Propan-l-ol

Functional group: OH (Alcoholic group)

Question 4.
Draw the structural formula of ethyne. How does the structure of alkynes differ from that of alkenes.
Answer:
Structural formula of ethyne
H —C ≡ C —H
In alkynes there are — C ≡ C — triple covalent bonds but in alkenes there areNew Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 8double covalent bonds.

Question 5.
Fill in the blanks with the correct words:
Alkenes are the_______ (analogous / homologous) series of_______ (saturated / unsaturated) hydrocarbons. They differ from alkanes due to the presence of_______ (double / single) bonds. Alkenes mainly undergo________ (addition / substitution) reactions.

Question 6.
Draw the structural formulae of the two isomers of Butane. Give the correct IUPAC name of each.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 9

2007

Question 1.
Give the IUPAC names of the compounds numbered (i) to (v).
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 10

Answer:

  1. Propyne
  2. Pentan-3-ol
  3. 2-Methylpropane
  4. Ethanoic acid
  5. 1,2-Dichloroethane

Question 2.
Copland complete the table which relates to three homologous series of Hydrocarbons :
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 11
Answer:

  1. Alkenes, Alkynes and Alkanes
  2. Double, Triple, Single
  3. Ethene, Ethyne, Methene
  4. Addition Substitution

Question 3.
Name the type of reaction by which X (compound of C and Br) can be prepared from ethane.
Answer:
By substitution reactions.

2008

Question 1.
The formation of 1, 2-dibromoethape from ethene and bromine is an example of :
A. Substitution
B. Dehydration
C. Dehydrohalogenation
D. Addition

Question 1.
Name the organic compound prepared by each of the following reactions :

  1. C2H5COONa + NaOH →
  2. CH3I + 2[H]-→
  3. C2H5Br + KOH (alcoholic soln.) →
  4. CaC2 + 2H2O →

Answer:
(1)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 12
Organic compound formed is Ethane.
(2)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 13
Organic compound formed is Methane
(3)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 14
Organic compound formed is Ethene.
(4)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 15
Organic compound formed is Acetylene.

Question 3.
Write the equation for the following :

  1. Calcium carbide and water
  2. Ethene and water (steam)
  3. Bromoethane and an aqueous solution of sodium

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 16

Question 4.
Distinguish between the saturated hydrocarbon ethane and the unsaturated hydrocarbon ethene by drawing their structural formulae.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 17

Question 5.
Which type of reaction i.e. addition or substitution is shown by ethane and ethene ?
Answer:
ethane→ Substitution, ethene → addition reaction

Question 6.
Write the equation for the complete combustion of ethane.
Answer:
2C2H6 + 7O2 → 4CO2 + 6H2O Δ

Question 7.
Name the alcohol, aldehyde and acid formed when ethane is oxidised.
Answer:
Alcohol obtained from ethane is ethyl alcohol [C2H5OH]. The aldehyde obtained from ethane is acetaldehyde [CH3 CHO]. The acid obtained from ethane is acetic acid [CH3COOH].

Question 8.
Why is pure acetic acid known as glacial acetic acid ?
Answer:
Pure acetic acid is known as glacial acetic acid because it freezes below 16.5°C to an icy mass (glacier).

Question 9.
What type of compound is formed by the reaction between acetic acid and an alcohol ?
Answer:
Ester is formed by the reaction between acid and an alcohol.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 18

Question 10.
By what type of reaction could a compound containing C, H and Cl – be obtained from ethyne ?
Answer:
Addition reaction
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 19

Question 11.
State the term for the reaction in which the hydrogen of an alkane is    replaced by chlorine.
Answer:
Substitution reaction.

2009

Question 1.
Which of the following statements is wrong about lkanes ?
(A) They are all saturated hydrocarbons.
(B) They can undergo addition as well as substitution reaction.
(C) They are almost non polar in nature.
(D) On complete combustion give out carbon dioxide and water.

Question 2.
Write balanced equation for : Acetic acid is warmed with ethanol in the presence of con. H2SO4.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 20
Question 3.
Find the odd one out in each case and explain your choice.

  1. C3H8 ,C5H10, C2H6, CH4
  2. Formic acid, Nitric acid, Acetic acid, Propanoic acid.

Answer:

  1. C5H10 [All others are alkane, this is an alkene]
  2. Nitric acid [This is the only inorganic acid rest all are organic acids]

Question 4.
Identity the substances ‘S’ based on the information given below:
The reddish brown liquid ‘S’ is dissolved in water. When ethyne gas is passed through it, ttirns colourless.
Answer:
Bromine solution

Question 5.
Fill in the blanks with the correct words from the brackets.

Generally ionic compounds exist in (i) ………………. (solid/liquid/gas) state. Melting and boiling points of covalent compounds are generally (ii) …………….. (low/high). The general formula for alkane is (iii)……………. (CnH 2n / CH 2n-2 t/CH2n+2). For alkynes the geperal formula is (iv)…… (CnH2n / CnH2n-2/CnH2n+2)

Question 6.
Give chemical equation for the following :

  1. The laboratory preparation of methane from sodium acetate.
  2. The reaction of one mole of ethene with one mole of chlorine gas.
  3. The preparation of ethyne from 1, 2 – dibromoethane.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 21

Question 7.
State how the following conversions can be carried out:

  1. Ethyl chloride to Ethyl alcohol.
  2. Ethyl chloride to Ethene.
  3. Ethene to Ethyl alcohol.
  4. Ethyl alcohol to Ethene.

Answer:

  1. By treating ethyl chloride with aqueous KOH.
  2. By heating ethyl chloride with alcoholic KOH.
  3. By passing ethene into concentrated H2SO4 at 80°C and high pressure or by hydrating of ethene.
  4. By heating ethyl alcohol with concentrated H2SO4 at 170°C.

Question 8.
Define isomerism. Give the IUPAC name of the isomer C4H10 which has a branched chain.
Answer:
(1) Isomerism : Organic compounds having the same  formula but different structural formulae are called isomers and this property is known as isomerism.
Example : n pentane & isopentane
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 22

2010

Question 1.
Select the correct answer

  1. The organic compound, which gives a red precipitate with ammoniacal cuprous chloride and undergoes an addition reaction –
    (A) Ethane
    (B) Ethene
    (C) Ethyne   
    (D) Ethanol
  2. The organic compound which when mixed with ethyle alcohol, [ethanol], makes it spurious.
    (A) Methanol
    (B) Methanoic acid
    (C) Methanal
    (D) Ethanoic acid

Question 2.
Draw the structural formula of—
1. Ethanoic acid
2. But-2-yne

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 23

Question 3.
Compound ‘X’ is bubbled through bromine dissolved in CCl4 and the product formed is CH2Br – CH2Br.

  1. Draw the structural of X and state what type of reaction X has undergone.
  2. State your observation for the above reaction.
  3. Name the compound formed when steam reacts with A in the presence of phosphoric acid.
  4. What is the procedure for converting the product of (b) (iii) back to X ?

Answer:

  1. CH2Br – CH2Br
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 24
    The above reaction is called addition reaction,
  2. The colour of bromine colour fades.
  3. New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 25
  4. Ethanol can be converted into ethene, by dehydrating it with cone, sulphuric acid.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 26

2011

Question 1.
Name a gaseous hydrocarbon commonly used for welding purposes.
Answer:
Acetylene

Question 2.
Give reasons for the following –

  1. almost 90% of all known compounds are organic in nature.
  2. it is dangerous to burn methane in an insufficient supply of air.

Answer:

  1. Because of ability of carbon to catenate i.e forms straight chain, branched chains or ring like compounds.
  2. Because carbon monoxide is produced in an insufficient supply of air. This gas is extremely poisonous for human beings as it cuts off the oxygen supply by forming carboxy haemoglobin in the blood.

Question 3.
Choose the correct answer –

  1. The functional group present in acetic acid is:
    (A) Ketonic C = O
    (B) Hydroxyl-OH
    (C) Aldehydic – CHO
    (D) Carboxyl – COOH
  2. Unsaturated hydrocarbons undergo :
    (A) a substitution reaction
    (B) an oxidation reaction
    (C) an addition reaction
    (D) none of the above
  3. The number of C – H bonds in ethane molecule are:
    (A) Four
    (B) Six
    (C) Eight
    (D) Ten

Question 4.
Select the correct answer the choices given :

  1. The catalyst used for conversion of ethene to ethane is commonly……… [nickel/iron/cobalt]
    Ans: Nickel
  2. Acetaldehyde when oxidized with acidified potassium dichromate, forms ………  [ester/ethanol/
    acetic acid]
    Ans: Acetic   acid
  3. Ethanoic acid reacts with ethanol in presence of cone. H2SO4, so as to form a compound and water. The chemical reaction which takes place is called…………[dehydration/hydrogenation/esterification]
    Ans: Esterification

Question 5.
Write balanced chemical equations for the following :

  1. Write the equation for the reaction taking place between 1, 2 – dibromoethane and ajcoholic potassium hydroxide.
  2. Monochloro ethan is hydrolysed with aqueous KOH.
  3. A mixture of sodalime and sodium acetate is heated.
  4. Ethanol under high pressure and low temperature is treated with acidified potassium dichromate.
  5. Water is added to calcium carbide.
  6. Ethanol reacts with sodium at room temperature.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 27

2012

Question 1.
State the observation : Bromine vapours are passed into a soln. of ethyne in carbon tetrachloride.
Answer:
Ethyne decolorizes the reddish brown colour of bromine solution.

Question 2.
From – Ethyne, ethanol, acetic acid, ethene, methane. Choose the one which relates to (i) to (iv).

  1. An unsaturated hydrocarbon used for welding purposes.
    Ans: Ethyne
  2. An organic compound whose functional group is carboxyl.
    Ans: Acetic acid
  3. A hydrocarbon which on catalytic hydrogennation gives a saturated hydrocarbon.
    Ans: Ethene
  4. An organic compound used as a thermometric liquid.
    Ans: Ethanol

Question 3.
(1) Why is pure acetic acid known as glacial acetic acid?
(2) Give a chemical equation for the reaction between ethyl alcohol and acetic acid.
Answer:

(1) Pure acetic acid freezes below 20°C to form a transparent solid which looks like ice and hence, it is called glacial acetic acid.
(2) New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 28

Question 4.
Rewrite the correct statement with the missing word/s. Ethyl alcohol is dehydrated by sulphuric acid at a temperature of about 170°C.
Answer:
Ethyl alcohol is dehydrated by concentrated sulphuric acid at a temperature of about 170°C.

Question 5.
Give the structural formula for the following :

  1. Methanoic acid
  2. Ethanal
  3. Ethyne
  4. Acetone
  5. 2-methyl propane.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 29

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 30


2013


Question 1.

Identify the gas evolved when : sodium propionate is heated with soda lime.
Answer:
Ethane gas

Question 2.
Give suitable chemical term for : A reaction in which hydrogen of an alkane is replaced by a halogen.
Answer:
Subsitution reaction

Question 3.
Give a chemical test to distinguish between : Ethene gas and ethane gas.
Answer:
To the given gas add few drops of bromine solution in carbon, tetra-chloride. In case of ethene gas, the reddish colour of bromine discharges. However, in case of ethane gas the reddish colour of bromine does not discharge.

Question 4.
Identify the statement that is incorrect about alkanes :
(A) They are hydrocarbons.
(B) There is single covalent bond between carbon and hydrogen
(C) They can undergo both substitution as well as addition reactions
(D) On complete combustion they produce carbon dioxide and water.

Question 5.
Give balanced equations for the laboratory preparations of:

  1. A saturated hydrocarbon from iodomethane.
  2. An unsaturated hydrocarbon from an alcohol.
  3. An unsaturated hydrocarbon from calcium carbide.
  4. An alcohol from ethyl bromide.

Answer:

  1. CH3I + 2H (from Zn/Cu couple) → CH4 + HI
  2. New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 31
  3. CaC2 + 2H2O → Ca(OH)2 + C2H2
  4. C2H5Br + KOH (aq) → C2H5OH + KBr

Question 6.
Give the structural formulae for :

  1. An isomer of n-butane.
  2. 2-propanol.
  3. Diethyl ether.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 32

Question 7.
Give reasons for :

  1. Methane does not undergo addition reactions, but ethene does.
  2. Ethyne is more reactive than ethane.
  3. Hydrocarbons are excellent fuels.

Answer:

  1. All the four covalent bonds between the carbon and hydrogen are fully shared. Thus the hydrogen atom can only be substituted by more reactive atoms or group of atoms. There is no scope of addition of reactive atoms in its molecule.
    However, in case of ethene there is a double bond between the two carbon atoms. These bonds are under strain and hence can be easily broken by more reactive atoms to form addition compounds which are saturated in nature.
  2. Ethyne has a triple covalent bond (—C = C —) between two carbon atoms, whereas ethene has a double covalent bond (—C = C —) between the two carbon atoms. So, the strain in the bounding of ethyne is far more than ethene. This accounts of the reactivity of ethyne as its bonds break more easily than that of ethene.
  3. All the constituents of hydrocarbon (carbon and hydrogen) are highly combustible and do not have any uncombustible content. So, hydrocarbons are excellent fuels.

2014

Question 1.
The I.U.P.A.C. name of acetylene is,
(A) propane
(B) propyne
(C) ethene
(D) ethyne.

Question 2.
Ethanol reacts with sodium to give………(sodium ethanoate, sodium ethoxide, sodium propanoate)
Answer:
sodium ethoxide

Question 3.
Give one word or phrase for – hydrocarbons containing a
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 33 functional group
Answer:
Hydrocarbons containing aNew Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 34functional group → Alkan one or Ketonic functional group.

Question 4.
Write balanced equation for preparation of

  1. ethane from sodium propionate.
  2. ethanol from monochloroethane and aq. sodium hydroxide.

Answer:

  1. Preparation of ethane from sodium propionate.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 35
  2. Preparation of ethanol from monochloroethane and aq. sodium hydroxide.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 36

Question 5.
Distinguish between : Ethane and ethene (using alkaline potassium permanganate solution)
Answer:
Ethane and ethene (using alkaline KMnO4)
Ethene decolourises the colour of alkaline KMnO4 but ethane does not

Question 6.
State the conditions required for :

  1. Catalytic hydrogenation of ethyne.
  2. Preparation of ethyne from ethylene dibromide.

Answer:

(1) One volume of ethyne gas is mixed with two volumes of hydrogen gas and passed over heated nickel at 300°C, when an addition reaction takes place with the formation of ethane gas.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 37

(2) Zinc dust is added to a mixture of 95% ethyl alcohol and 5% of ethylene dibromide. The mixture is gently warmed when ethyne gas is liberated.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 38

Question 7.
Write structural formula of:
(1) Ethanol
(2) 1-propanal.
(3) ethanoic acid.
(4) 1, 2, dichloroethane.
Answer:
Give the structural formula:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 39

Question 8.
Match A and B with (i) and (ii) :
A: alkynes    (1) CnH2n+2
B: alkane      (2) CnH2n-2 

Answer:

A: alkynes    (2) CnH2n-2
B: alkane      (1) CnH2n+2 

2015

Question 1.
Select from the list — Ammonia, ethane, hydrogen chloride, hydrogen sulphide, ethyne

  1. The gas is used for welding purposes.
    Ans: Ethyne
  2. This gas is also a saturated hydrocarbon.
    Ans: Ethane

Question 2.
State which of the following statements does not describe the property of alkenes :
(A) They are unsaturated hydrocarbons
(B) They decolourise bromine water
(C) They can undergo addition as well as substitution reactions
(D) They undergo combustion with oxygen forming carbon dioxide and water.
Ans. (C) They can undergo addition and substitution reactions. Alkenes do not undergo substitution reaction.

Question 3.
State one appropriate observation when : The gaseous product obtained by dehydration of ethyl alcohol is passed through bromine water.
Answer:
The reddish brown colour of bromine solution gets decolourised.

Question 4.
Give balanced chemical equations for the following con­versions:

  1. Ethanoic acid to ethyl ethanoate.
  2. Calcium carbide to ethyne.
  3. Sodium ethanoate to methane.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 40

Question 5.
Using their structural formulae identify the functional group by circling them:
(1) Dimethyl ether.
(2) Propanone
Answer:
Dimethyl ether.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 41
Propanone
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 42

Question 6.
Name the following :

  1. Process by which ethane is obtained from ethene.
  2. A hydrocarbon which contributes towards the greenhouse
  3. Distinctive reaction that takes place when ethanol is treated with acetic acid.
  4. The property of element by virtue of which atoms of the element can link to each other in the form of a long chain or ring structure.
  5. Reaction when an alkyl halide is treated with alcoholic potassium hydroxide.

Answer:

  1. Hydrogenation (addition)
  2. Methane
  3. Esterification
  4. Catenation
  5. Dehydrohalogenation

2016

Question 1.
Fill in the blanks : Conversion of ethene to ethane is an example of………… (hydration / hydrogenation).
Answer:
Conversion of ethene to ethane is an example of hydrogenation.

Question 2.
Write balanced chemical equations for : Preparation of ethanol from ethyl chloride.
Answer:
C2H5C1 + NaOH(aq) C2H5OH + NaCl

Question 3.
Identify the term/substance in each of the following :

  1. The catalyst used in the conversion of ethyne to ethane.
  2. The type of reactions alkenes undergo.

Answer:

  1. Nickel or platinum or palladium.
  2. Addition reactions.

Question 4.
Write the IUPAC names of:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 43
Answer:
(a) Propene
(b) 2-butyne
(c) ethanal

Question 5.
Write a balanced chemical for :

  1. Burning of ethane in plentiful supply of air.
  2. Action of water on calcium carbide.
  3. Heating of Ethanol at 170°C in the presence of cone, sulphuric acid.

Answer:

(1) C2H6 + 7/2 O2 → 2CO2 + 3H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
(2) CaC2 + 2H2O → Ca(OH)2 + C2H2
(3) New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 44

Question 6.
Give the structural formulae of:
(1) 2-methyl propane
(2) Ethanoic acid
(3) Butan – 2 – ol
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 45
Question 7.
Compound A is bubbled through bromine dissolved in carbon tetrachioride is as follows :
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 46

(1) Draw the structure if A.
(2) State your observation during this reaction.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 47

2017

Question 1.
Fill in the blanks from the choices given in brackets – The compound formed when ethene reacts with hydrogen is……….. [CH4, C2H6, C3H8]
Answer:
The compound formed when ethene reacts with hydrogen is C2H6.

Question 2.
Choose the correct answer from the options given – If the molecular formula of an organic compound is C10H18 it is –
(A) Alkene
(B) Alkane
(C) Alkyne
(D)  Not a hydrocarbon

Question 3..
Identify the substance underlined – An organic compound containing – COOH functional group.
Answer:
Ethanoic acid [CH3—COOH]

Question 4.
Write the balanced chemical equation for – Preparation of methane from iodomethane
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 48

Question 5.
Identify the term or substance based on the descriptions given below:

  1. Ice like crystals formed on cooling an organic acid sufficiently.
  2. Hydrocarbon containing a triple bond used for welding purpose.
  3. The property by virtue of which the compound has the same molecular formula but different structural formulae.
  4. The compound formed where two alkyl groups are linked byNew Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 49group.

Answer:

  1. Glacial acetic acid
  2. Ethyne or acetylene
  3. Isomerism
  4.  Ketone or Alkanone

Question 6.
Give a balanced chemical equation for each of the following –

  1. Preparation of ethane from sodium propionate.
  2. Action of alcoholic KOH on bromoethane.

Ans:

  1. C2H5COONa + NaOH → C2H6 + Na2CO3
  2. CH3Br + KOH → CH3OH + KBr

Question 7.
State one relevant observation for the following reaction – Addition of ethyl alcohol to acetic acid in the presence of concentrated sulphuric acid.
Answer:
On warming the mixture gives fruity smell.

Question 8.
Draw’ the structure formula for each of the following –

  1. 2, 3 – dimethyl butane
  2. Diethyl ether
  3. Propanoic acid

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 50

ADDITIONAL QUESTIONS


Question 1.
Explain the term ‘Organic Chemistry ’. State the ‘Natural sources ’ and ‘Importance’ of organic compounds.
Answer:

  1. Organic Chemistry-It is the chemistry of specific carbon compounds except – oxides, carbonates, bicarbonates and metallic carbides.
  2. Plants, Animals, Petroleum, dyes and drugs are all natural sources.
  3. Compounds of organic origin are : Food – carbohydrates, vitamins Dyes-azodyes Clothing – cotton, silk and wool Fuels – petrol Medicines – penicillin Explosives – trinitrotoluene.

Question 2.
Explain the ‘unique nature of carbon atom’ with specific reference and meaning to —
(a) ‘Tetravalency’ — leading to formation of single, double and triple bonds
(b) ‘Catenation’ — leading to formation of straight chain, branch chain and cyclic compounds.
Answer:
Some unique properties shown by carbon atom are :
(a) Tetravalency
(b) Catenation
(c) Ability to form multiple bonds.
(a) Tetravalency : Atomic number of carbon is 6. Its electronic configuration is 2, 4. Therefore, it has four electrons in its valence shell. Carbon atom can neither lose nor gain electrons to complete its octet (not possible from energy point of view). Therefore, carbon atom completes its octet by sharing four electrons with other atoms, i.e., it can form four covalent bonds, called its tetracovalency.
For example:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 51

(b) Catenation: The property by virtue of which a large number of atoms of the same element get linked together through single or multiple covalent bonds, forming straight or branched chains and rings of different sizes, is called catenation. Carbon shows catenation to the maximum extent due to strong carbon-carbon bonds and its tetracovalency.

In this process of catenation, carbon atoms form straight or branched chains and cyclic rings of various sizes and can involve single, double or triple covalent bonds.

Question 3.
State reasons for ‘Justification of a separate branch’ for ‘Organic Chemistry.
Answer:
This is due to the following reasons:

  1. The number of known organic compounds is very large as compared to the number of known inorganic compounds.
  2. Organic compounds involve only a few elements (C, H, O, N, S, P, F, Cl, Br, I etc.), whereas inorganic compounds involve all the known elements.
  3. Organic compounds have complex nature and have high molecular mass.
  4. Organic compounds involve covalent bonds whereas inorganic compounds involve electrovalent bonds.
  5. Organic compounds show isomerism whereas inorganic compounds do not show isomerism.
  6. The properties of organic compounds are different from inorganic compounds.
    All these facts convince us to study organic chemistry as a separate branch of chemistry.

Question 4.
State five differences between the characteristics of organic and inorganic compounds. State how organic compounds are classified.
Answer:
(a) Characteristics of organic compounds :

  1. These are made up of only a few elements C, H, O, N, S,X(C1, Br,l)
  2. These involve covalent bonds.
  3. These are generally gases or liquids
  4. They have low melting and boiling points.
  5. They are combustible.
  6. They show molecular reations.
  7. They show isomerism.
  8. These are non-conductors of electrocity.
  9. There are generally insoluble in water but soluble in organic solvents.

Characteristics of inorganic compounds :

  1. These are made up of all the known elements.
  2. These involve ionic bonds.
  3. These are generally solids.
  4. They have high melting and boiling points.
  5. They are non-combustible.
  6. They show ionic reactions.
  7. They don’t show isomerism.
  8. These are generally good conductors of electricity.
  9. These are generally soluble in water but insoluble in organic solvents.

(b) Classification of Organic Compounds Aliphatic – Open Chain Compounds
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 52

Question 5.
Explain the term ‘Homologous series’. State the general characteristics of members of the series with special reference to molecular mass or molecular formula.
Answer:
Homologous series is a series of organic compounds, that are grouped into a smaller number of series of compound.
General Characteristics of homologous series :

  1. The members of a series have same functional group.
  2. Two consecutive members of a homologous series differs each other in their composition
    by – CH2unit
    Example :
    Alcohol (-OH)
    CH3 – OH, CH3 – CH2 – OH, CH3 – CH2 – CH2 – OH
  3. The members of a homologous series can be represented by same general formula.
    Example :
    Alcohol-CnH2n+1OH
    Aldehyde — CnH2n+1|CHO
    Carboyxlic acid — CnH2n+1COOH
  4. The members of a particular homologous series have almost same chemical properties due to presence of same functional group.
  5. The physical properties (like solubility, melting point, boiling point, state) of members of a homologous series either gradually increase or decrease with increase in molecular mass.
  6. The members of a homologous series can be prepared by same or common general method of preparation.
  7. The first member of homologous series generally shows certain different chemical behavior than other members of the series.

Question 6.
Differentiate between — ‘Molecular formula’ and ‘Structural formula’ — of an organic compound. Write the ‘condensed structural formula and ‘branched structural formula’ of ethene.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 53

Question 7.
State what are ‘Alkyl groups ’. State the alkyl group of the parent alkane — methane and ethane.
Answer:
Alkyl Group : It is obtained by removing one hydrogen atom from a molecule of an alkane.
Methane : Methyl (Alkyl group)
Ethane : Ethyl (Alkyl group)

Question 8.
State what are ‘Functional groups’. Name the following functional groups —
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 54
X = -F, -Cl, -Br, -I ; -C=O; -C-O-C
Answer:
Functional Groups : An atom, radical or bond which defines the structure of an organic compound and give if its characteristic properties
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 55

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 56
Question 9.

Explain the terms — ‘Isomers’ and ‘Isomerism’. State the ‘Characteristics of isomers’ with reference to —
Properties of isomers ; Number of isomers with relation to carbon atoms in the isomer.
Differentiate between — ‘Chain isomerism’ and ‘Position isomerism ’ – with suitable examples.
Answer:
Two or more compounds having the same molecular formula but different physical and chemical properties are called isomers and this phenomenon is known as isomerism.
Isomers have the same number of atoms of each element in them and the same atomic weight but differ in other properties. For example, there are two compounds with the molecular formula C2H6O. One is ethanol (ethyl alcohol), CH3CH2OH, a colorless liquid alcohol; the other is dimethyl ether, CH3OCH3, a colorless gaseous ether.
Alkanes with more than three carbon atoms form isomers. The various isomers differ in the framework of the carbon chains.
Differentiate between — ‘Chain isomerism’ and ‘Position isomerism’

Chain isomers: Compounds having same molecular formula with difference in carbon chain pattern like linear or branch are called chain isomers. 1-Pentyne is chain isomer for 3-methyl Butyne.
CH3 – CH2 – CH2 – C = CH and CH3– C(CH3) – C = CH
Position isomers: Compounds having same molecular formula ^ with difference in position of the functional group are called position isomers. 1-Butyne and 2-Butyne are position isomers.
CH3– CH2– C=CH and CH3 – C = C – CH3.

Question 10.
Explain the term – ‘Nomenclature’. State its need with reference to organic compounds. State the basic rules of Nomenclature by the trivial system – with suitable examples. Explain the longest chain rule and the smallest number for functional groups rule of Nomenclature by the IUPAC system – with suitable examples.
Answer:
(a) Nomenclature :
Nomenclature is the system of assignment of names to organic compounds.
Need for Nomenclature : Very large number of organic compounds with varying molecular structure need a systematic method of nomenclature. Further many a times same molecular formula represents two or more compounds (isomerism).
(b) Nomenclature by Trivial System:
In this method, name of an organic compound is derived from its;

  1. Source (e.g., benzoic acid is obtained by distillation from gum benzoin, fructose or fruit sugar from fruits etc).
  2. Latin or Greek origin (e.g., formic acid, HCOOH is present in sting of red ants, formicus in Latin means an ant).
  3. Properties (e.g., palmitic acid is an acid derived from palm oil etc).

(c) Longest Chain Rule :

  1. In the nomenclature of alkanes, the longest continuous chain if C-atoms is selected. For this, alkyl groups, if present, are written in the expanded form.
    For example,
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 57
  2. Smallest Number for Substituent : Once the principal chain is selected, it is numbered in such a say that the substituent gets the lowest number.
    For example,
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 58

Question 11.
Explain the term – ‘Hydrocarbons’. State the two main groups of hydrocarbons with examples. Draw a chart differentiating — ‘Alkanes, Alkenes andAlkynes’ — with respect to:

  1. General formula
  2. Characteristic bond
  3. IUPAC and the common name of the first three members and condensed/branched/electronic structural formula of each
  4. Availability of electrons
  5. Reactivity
  6. Characteristic reaction.

Answer:

Hydrocarbons — They are aliphatic open chain organic compounds containing carbon and hydrogen only.
Molecular formula is CxHy where X and Y are whole numbers.

  1. Saturated hydrocarbons — Homologous series of alkanes.
  2. Unsaturated hydrocarbons — Series of alkynes and alkenes.
    • General formula :
      Alkanes — CnH2n+2
      Alkenes — CnH2n
      Alkynes — CnH2n_2
    • Characteristic bond
      Alkanes → C – C < single bond
      Alkenes → C = C < double bond
      Alkynes → C = C < triple bond
    • IUPAC Name Condensed/branched/electronic structural formula of each
      New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 59..New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 60New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 61
    • Availability of electrons
      Alkanes — Not available
      Alkenes — Available Alkynes — Available
    • Reactivity
      Alkanes — Less reactive
      Alkenes — More reactive
      Alkynes — Most reactive
    • Characteristic Reaction
      Alkanes — Substitution reaction
      Alkenes — Addition reaction
      Alkynes — Addition reaction

Question 12.
Draw the structural formula of each of the following :

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 62

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 63

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 64

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 65
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 66

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 67

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 68

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 69

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 70

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 71

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 72

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 73

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 74

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 75

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 76

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 77

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 78New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 79
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 80

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 81

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 82

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 83

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 84
Question 13.
Give the IUPAC name of the compounds numbered (I) to (y).

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 85
Answer:

  1. Methyl butanol, 2-Methyl-1-butanol
  2. 2, 2-dimethyl propanol
  3. 2-Bromocyclo pentan- I -ol
  4. 3-Methylbutanal
  5. 3-Methyl-2-butanone

Question 14.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 86

Answer:

(a) CH3COONa + NaOH → CH4 + Na2CO3
(b) CH3I + 2|H| → CH4 + HI
(c) C2H5COONa + NaOH → C2H6 + NaCO3
(d) C2H5Br + 2(H) → C2H6 + HBr

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 87

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 88
Question 15.
Give equations for the conversions of – Methane, Ethane, Ethene, Ethyne, Methanol, Ethanol and Ethanoic Acid.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 89
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 90
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 91
When methane is burnt in excess of air or oxygen with pale blue flame it gives carbon dioxide gas, water and heat energy. This reaction is complete oxidiation reaction.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 92
This reaction is carried out in a copper tube. Cu acts as a catalyst. In this reaction, methane is oxidised to methanol or methyl alcohol. In this reaction, methane is heated up to a temp, of 200°C under a pressure of 120 atm.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 93
Methane can also be converted into methanol by controlled oxidation of methane in the presence of acidified K2Cr2O7.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 94
In this reaction, methane is oxidised in a copper tube methane is heated upto a tem. of 475 k under a pressure of 120 atm. Copper tube acts as a catalyst. This is carried out as catalystic oxidation.

(d) Conversion of methane to Methanal
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 95
This reaction involves the catalytic oxidation. In this reaction, methane is heated with catalyst molybdenum oxide (MoO) it a temp, of 350 – 500°C, methanol is formed.

(e) Conversion of Methane to Ethyne (C2H2)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 96
When methane is heated to about 1500°C in an electric arc and then suddenly cooled, the product is C2H2and Hydrogen.
Conversion of Ethane (C2H6) to
(a) Hexachloro ethane (C2Cl6)

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 97
C2H4Cl2 + Cl2 → C2H4Cl2 + HCl
C2 H4Cl2 + Cl2→ C2H3Cl3 + HCl
C2 H3Cl3 + Cl2 → C2H2Cl4 + HCl
C2 H2Cl4 + Cl2→ C2HCl5 + HCl
C2HCl5 + Cl2 → C2Cl6 + HCl
This reaction is a substitution reaction.

(b) Carbon dioxide (CO2)

2C2H6 + 7O2 (excess) →4CO2 + 6H2O

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 98

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 99

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 100
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 101
(a) Ethene
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 102

(b) 1,1,2,2, tetrachloroethane

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 103
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 104
(c) 1,1,2,2 tetrabromoethane

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 105
(d) 1,2 diiodoethane
C2H2+I2 →C2H2I2                           1,2 diiodoethene

(e) 1,1,Dibromoethane

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 106
(f) Copper acetylide / Silver acetylide

HC ≡ CH+ 2CuCl + 2NH2OH →
Cu – C ≡ C – Cu + 2NH2Cl + 2H2O  Copper acetylide
HC ≡ CH + 2AgNO3 + 2NH4OH →
Ag – C= C – Ag + 2NH2NO4 + 2H2O Silver acetylide
Conversion of Ethanol to :
(a) Carbon dioxide

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 107

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 108

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 109
Question 16.
Give reasons for

  1. alkanes are said to be saturated organic compounds
  2. alkenes are known as olefins
  3. alkenes are more reactive than alkanes
  4. ethanoic acid is known as an aliphatic monocarboxylic acid.

Answer:

  1. Alkanes do not undergo addition reactions and that is why they are called saturated hydrocarbons or saturated organic compounds. In alkanes all the four valencies of carbon atom are fully satisfied by forming single covalent bonds.
  2. Alkenes are called olefins because alkenes on treatment with halogens form oily products. (Latin: oleum = oil, ficare = to make)
  3. Due to the presence of C = C (carbon – carbon double bond) alkenes are more reactive than alkanes.
  4. Ethanoic acid (CH3 – COOH) contains only one – COOH group (carboxylic acid group) that is why it is called a monocarboxylic acid. As ethanoic acid does not contain a benzene right it is an alphatic monocarboxylic acid.

Question 17.
Explain the terms –

  1. Denaturated alcohol
  2. Glacial acetic acid
  3. Esterification

Answer:

  1. Denaturated alcohol – Ethyl alcohol containing pyridine or copper sulphate is termed – denaturated alcohol. It is used for – industrial applications only and hence made undrinkable.
  2. Glacial acetic acid – Anhydrous acetic acid on cooling below 16.5°C crystallizes out in the pure form, forming a crystalline mass resembling ice. Hence pure acetic acid is called glacial acetic acid.
  3. Esterification – It is known as condensation of an alcohol with an acid. Acetic acid on heating with an alcohol and dehydrating agent [cone. H2SO2] gives an ester – ethyl acetate.

Question 18.
Give a chemical test for to distinguish between

  1. Ethane, ethene and ethyne
  2. Ethanol and ethanoic acid.

Answer:

  1. Tests to distinguish between ethane, ethene and ethyne
    • Test — Br2 water test: Pass the gas through Br2 water,
      Ethane : Brown colour of Br2 water is not discharged.
      Ethene : Brown colour of Br2 water is discharged,
      Ethyne : Brown colour of Br2 water is discharged.
    • Test — Baeyer’s reagent : Pass the gas through
      Baeyer’s reagent (alkaline solution of KMnO4).
      Ethane : Purple colour of Baeyer’s reagent is not discharged.
      Ethene : Purple colour of Baeyers reagent is discharged.
      Ethyne : Purple colour of Baeyers reagent is discharged.
    • Test : Pass the gas through ammoniacal cuprous chloride solution.
      Ethane : No ppt.
      Ethene : No ppt.
      Ethyne : Red ppt. of copper acetylide is formed.
    • Test : Pass the gas through ammonical silver nitrate solution.
      Ethane : No ppt.
      Ethene : No ppt.
      Ethyne : White ppt. of silver acetylide is formed.
  2. Tests to distinguish between ethanol and ethanoic acid
    1. Test: Add a few drops of blue litmus solution to the given liquid.
      Ethanol: No change in colour.
      Ethanoic acid : Blue litmus turns red.
    2. Test: Add a pinch of sodium carbonate to the given liquid. Ethanol: No action.
      Ethanoic acid : Brisk effervescence with the evolution of C02.

Question 19.
Give the main uses of –

  1. Methane
  2. Ethane
  3. Ethene
  4. Ethyne
  5. Ethanol
  6. Ethanoic acid.

Answer:
The main uses of:
(1) Methane and

(2) Ethane –

(a) Illuminant and domestic fuel: In the form of natural gas or gobar gas. [Hydrocarbons – have high calorific value. They are easily combustible and the reaction is exothermic – releasing heat energy. Hence they are excellent fuels]
(b) In manufacture of chemicals : Used as :

  1. Chloroform : Solvent for rubber, waxes. As an anaesthesia.
  2. Carbon black : A black pigment in shoe polishes, printers ink etc.
  3. Formaldehyde : An antiseptic, preservative for biological specimens.
  4. Methanol : Solvent for varnishes, anti-freeze for automobiles.
  5. Ethanol: Solvent for resins, a low freezing liquid in thermometers.

(3)  Ethene –

(a) Production of oxy-ethylene torch : For welding purposes and cutting metals.
(b) Ripening of green fruits : Artificial ripening and preservation of fruits.
(c) Catalytic hydrogenation: Used in hardening of oils.
(d) It is also used in manufacturing of :

  1. Synthetic chemicals : Ethylene glycol [anti-freeze], di-ethyl ether [solvent], ethylene oxide [fumigant], mustard gas [chemical warfare],
  2. Polymers : Polyetheylene, polyvinyl chloride [P.V.C.]- used in packaging, insulators, containers, rain coats etc.

(4) Ethyne –

(a) It is used for producing oxy-acetylene flame for welding and cutting purposes as it produces temperature as high as 3500°C.
(b) It is used as an illuminant in oxyacetylene lamp.
(c) It is used in the manufacture of solvent like westron (C2H2C14) and westrosol (CHCl = CC2).

(5) Ethanol – Main Uses of Ethanol

(a) As a solvent – For gums and resins
(b) In thermometers and spirit levels – Low freezing mobile liquid, [freezing point – 114.1°C].
(c) In manufacture of chemicals – Acetaldehyde [dyes], acetic acid [manufacture of vinegar], chloroform [antiseptic] diethyl ether [anaeshetic].

(6) Ethanoic acid –
Uses :

(a) It is used as a solvent for many organic reactions.
(b) It is used as vinegar for preparing pickles etc.
(c) It is used for preparing various organic compounds like acetone, acetic anhydride ester etc.
(d) It is used as cogulating agent in rubber industries.
(e) It is used for making perfumes and medicines.

UNIT TEST PAPER 8 — ORGANIC CHEMISTRY

Question 1.
Draw the branched structural formula of the following organic compounds whose IUPAC names are given below.

  1. Pent-l-ene
  2. But-2-yne
  3. 3-methyl pentane
  4. 2-methyl-prop-l-ene
  5. Pentane-3-ol
  6. 1, 1, 2, 2 tetrabromoethane
  7. 2-methyl butan -2-ol
  8. 2, 2 dimethylpropan-l-ol
  9. 2, 2 dimethyl propane
  10. 2-bromo-4-chloro pentane

Ans.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 110

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 111
Question 2.
Select the correct answer from the choice in brackets.

1. The vapour density of the fifth member of the homologous series of alkanes. [22 / 36 / 29]
2. The isomer of pentane which has ‘1’ C atom attached to ‘4’ other C atoms [n – / iso- / neo-] pentane.
3. The IUPAC name of the product of reaction of ethylene with hydrogen bromide, [ethyl bromide / bromoethane / dibromoethane]
4. The IUPAC name of methyl acetylene. [1-butyne / propyne / ethyne]
5. The functional group in ethanoic acid, [aldehydic / carboxyl / hydroxyl]
Ans:

  1. 36
  2. neo-pentane
  3. bromoethane
  4. propyne
  5. carboxyl

Question 3.
Give balanced equations for the following conversions.New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 112
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 113
Question 4.
Select from the letters A to G the correct answer corresponding to the statements from 1 to 5 :
A :Ammoniacal CuCl2,
B : Trichloromethane,
C : Trichloroethane,
D : Bromine soln.,
E : Aqueous KOH
F : Ethene,
G : Sodalime,
H : Ethanol,
I : Ethyne.

  1. The organic compound which forms carbon tetrachloride on reaction with chlorine.
  2. The reagent which can distinguish between ethene and ethyne.
  3. The substance which reacts with bromoethane to give ethanol.
  4. The substance which gives bromoethane on reaction with hydrogen bromide.
  5. The substance which reacts with acetic acid to give CH3COOC2H5

Answer:

(1) -B
Explination

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 114
(2) – A
Explanation: [Only ethyne gives red ppt. of dicopper acetylide with ammoniacal Cu2Cl2 or CuCl]

(3) -E
Explination

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 115

(4) -F
Eplination
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 116

(5) -H
Explanation
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 117

Question 5.
Give balanced equations for the following conversions.

  1. An alkyne to an alkene.
  2. An alkene to an alkane.
  3. An alkane to an alcohol.
  4. An alcohol to an alkene.
  5. A carboxylic acid to an ammonium salt

New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 118

Question 6.
Give reasons for the following :

Question 6(1).
Concentrated sulphuric acid maybe added during esterification of acetic acid.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 8 Organic Chemistry 119

Cone. H2SO4, a strong dehydrating agent helps in the removal of water thus shifting the equilibrium in the forward direction, resulting the formation of more of ethyl acetate (ester).

Question 6(2).
Isomers belonging to the same homologous series may differ in physical properties but not in chemical properties.
Answer:
Isomers of the same homologous series have the same functional group (if any) and as such have similar chemical properties. As isomeres are different compounds they differ in one or more of their physical properties.

Question 6(3).
A given organic compound can be assigned only one name on the basis of the IUPAC system.
Answer:
This statement is not correct.Correct statement is : An organic compound may have more than one IUPAC name  (out of all these one is a preferred IUPAC name) but two compounds cannot have the same IUPAC name because thismay lead to confusion.

Question 6(4).
Substitution reactions are characteristic reactions of saturated organic compounds only.
Answer:
Addition reactions are not possible in case of saturated organic compounds. Saturated organic           compounds can only undergo substitution reactions.

Question 6(5).
Acetic acid is considered an aliphatic monocarboxylic acid.
Answer:
Acetic acid or ethanoic acid, CH3COOH has one carboxylic acid group (—COOH). Hence it is a monocarboxylic acid. As it has no benzene ring in it, it is  not aromatic and hence it is an aliphatic monocarboxylic acid.

 

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New Simplified Chemistry Class 10 ICSE Solutions – Practical Chemistry

New Simplified Chemistry Class 10 ICSE Solutions – Practical Chemistry

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These Solutions are part of New Simplified Middle School Chemistry Class 10 ICSE Solutions Here we have given New Simplified Chemistry Class 10 ICSE Solutions – Practical Chemistry

2003

Question 1.
State the colour of the residue obtained on cooling when the following carbonates are heated :

  1. zinc carbonate
  2. lead carbonate
  3. copper carbonate.

Answer:

  1. In case of zinc carbonate, residue is yellow when hot and White when cold
    New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 4
  2. In case of lead carbonate, residue is reddish-brown when hot and yellow when cold
    New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 4.1
  3. In case of Copper carbonate, residue is black when cold.
    New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 4.2

2004

Question 1.
Sodium hydroxide solution is added first in a small quantity, then in excess to the aqueous salt solutions of copper (II) sulphate, zinc nitrate, lead nitrate, calciumchloride and iron (III) sulphate. For each of the aqueous salt solutions, state —

(a) the colour of the precipitate when NaOH is added in a small quantity ;
(b) the nature of precipitate (i.e. soluble or insoluble) when NaOH is added in excess.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 5

2005

Question 1.
The questions below refer to the following salt solutions listed A to F : A : Copper nitrate B : Iron (II) sulphate C : Iron (HI) chloride D : Lead nitrate E : Magnesium sulphate F : Zinc chloride.

  1. Which two solutions will give a white precipitate when treated with dilute hydrochloric acid followed by barium chloride solution, (i.e. white ppt. insoluble in dil. HCl)
  2. Which two solutions will give a white ppt. when treated with dil. HNO3 and AgNO3 soln.
  3. Which soln. will give a white ppt. when either dil. HCl or dil. H2SO4 is added to it.
  4. Which soin. becomes a deep/inky blue colour when excess of ammonium hydroxide is added to it.
  5. Which solution gives a white precipitate with excess ammonium hydroxide solution.

Answer:

  1. B and E [Iron(II) Sulphate and Magnesium sulphate]
  2. C and F [Iron(II) chloride and Zinc chloride]
  3. D [Lead nitrate]
  4. A [Copper nitrate]
  5. D [Lead nitrate]

2006

Question 1.
From the list of substances given—Ammonium sulphate, Lead carbonate, Chlorine, Copper nitrate, Ferrous sulphate — State : A substance that turns moist starch iodide paper blue.
Answer:
Chlorine

Question 2.
State what is observed when excess of ammonia passed through an aqueous solution of lead nitrate.
Answer:
When ammonia is passed through an aq. solution of lead nitrate, chalky white precipitate of lead hydroxide is formed which is insoluble in excess of ammonia.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 6

Question 3.
Give one test each to distinguish between the following pairs of chemicals solutions

  1. Zn(NO3)2 and Ca(NO3)2
  2. NaNO3 and NaCI
  3. Iron [III] chloride and copper chloride.

Answer:

  1. Add NaOH solution in excess to the two solutions. The one in which white ppt. initially formed dissolves in excess of NaOH solution is Zn(NO3)2 solution and the other is Ca(NO3)2 solution.
  2. Add freshly prepared ferrous sulphate solution to the two solutions. Then by the side of the test tube, pour cone, sulphuric acid (H2SO4) to each slowly. The one in which brown ring appears is sodium nitrate solution while the other is sodium chloride sol.
  3. Add NaOH solution to both the solutions. The one which give a reddish brown ppt. is Iron(II) chloride solution and the one which gives blue ppt. is copper chloride solution.

Question 4.
Give a reason why carbon dioxide and sulphur dioxide cannot be distinguished by using lime water.
Answer:
Because both turns lime water milky.

2007

Question 1.
Salts A, B, C, D and E undergo reaction (i) to (v) respectively. Identify the anion present in these salts on the basis of these reactions.

  1. When AgNO3 solution is added to a soln. of A, a white precipitate, insoluble in dilute nitric acid, is formed.
  2. Addition of dil. HCl to B produces a gas which turns lead acetate paper black.
  3. When a freshly prepared solution of FeSO4 is added to a soln. of C and cone. H2SO4 is gently poured from the side of the test-tube, a brown ring is formed.
  4. When dil. H2SO4 is added to D a gas is produced which turns acidified K2Cr2O7 soln. from orange to green.
  5. Addition of dil. HCI to E produced an effervescence. The gas produced turns limewater milky but does not effect acidified K2Cr2O7 soln.

Answer:

  1. Chloride Cl1-
  2. Sulphide S2-
  3. Nitrate NO32-
  4. Sulphite SO32-
  5. Carbonate CO32-

Question 2.
How will the addition of barium chloride soln. help to distinguish between dil. HCl and dil. H2SO4.
Answer:
Barium chloride does not react with dil HCl but with dil H2SO4 gives a white ppt. of barium sulphate.

2008

Question 1.
The salt which, in solution gives a pale green ppt. with sodium hydroxide solution and a white precipitate with barium chloride solution is :

A : Iron (III) sulphate
B : Iron (II) sulphate
C : Iron (II) chloride
D : Iron (III) chloride

Answer:

B : Iron (II) sulphate

2009

Question 1.
Carbon dioxide and sulphur dioxide gas can be distinguished by using :

(A) Moist blue litmus paper
(B) Lime water
(C) Acidified potassium dichromate paper
(D) None of the above.

Answer:

(C) Acidified potassium dichromate paper

Question 2.
Identity the substances ‘R’ based on the information given below :
The pale green solid ‘R’ turns reddish brown on heating. Its aqueous solution gives a white precipitate with barium chloride solution. The precipitate is insoluble in mineral acids.
Answer:
Ferrous sulphate (Fe2SO4).

Question 3.
Give one chemical test to distinguish between the following pairs of compounds.

  1. ZnSO4 and ZnCl2
  2. FeCl2 and FeCl3
  3. Calcium nitrate soln. and Calcium chloride soln.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 8
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 9

2010

Question 1.
Select the correct answer from A, B, C, D and E –

(A) Nitroso Iron (II) sulphate
(B) Iron (III) chloride
(C) Chromium sulphate
(D) Lead (II) chloride
(E) Sodium chloride.

The compound which is responsible for the green colour formed when SO2 is bubbled through acidified potassium dichromate solution.
Answer:

(C) Chromium sulphate

Question 2.
State your observation –

  1. A piece of moist blue litmus paper
  2. Paper soaked in potassium permanganate solution- is introduced into a gas jar of sulphur dioxide.

Answer:

  1. Moist blue litmus turns red and finally colourless as SO2 act as an acidic gas and then a bleaching agent.
  2. The pink colour of potassium permanganate paper turns colourless because of bleaching property of sulphur dioxide.

Question 3.
Write the equation for the reaction of magnesium sulphate solution with barium chloride solution.
Answer:
MgSO4 + BaCl2 → BaSO4 + MgCl2

2011

Question 1.
Choose from the list of substances – Acetylence gas, aqua fortis, coke, brass, barium chloride, bronze, platinum.
An aqueous salt solution used for testing sulphate radical.
Answer:
Barium chloride.

2012

Question 1.
Name the gas which turns acidified potassium dichromate clear green.
Answer:
Sulphur dioxide gas (SOs)

Question 2.
Identify the anion present in the following compounds :

  1. Compound X on heating with copper turnings and concentrated sulphuric acid liberates a reddish brown gas.
  2. When a solution of compound Y is treated with silver nitrate solution a white precipitate is obtained which is soluble in excess of ammonium hydroxide solution.
  3. Compound Z which on reacting with dilute sulphuric acid liberates a gas which turns lime waer milky, but the gas has no effect on acidified potassium dichromate solution.
  4. Compound L on reacting with Barium chloride solution gives a white precipitate insoluble in dilute hydrochloric acid or dilute nitric acid.

Answer:

  1. Nitrate ion, NO3
  2. Chloride ion, Cl
  3. Carbonate ion, CO32-
  4. Sulphate ion, SO42-

Question 3.
State one chemical test between each of the following pairs :

  1. Sodium carbonate and Sodium sulphite
  2. Ferrous nitrate and Lead nitrate
  3. Manganese dioxide and Copper(II) oxide

Answer:

  1. Treat each of the compound with dilute sulphuric acid. In case of sodium carbonate a colourless and odourless gas is evolved. In case of sodium sulphite a colourless gas evolved which has a choking smell and causes coughing.
  2. To each of the aqueous solutions of compounds add aqueous sodium hydroxide solution. A dirty green precipitate is formed in case of ferrous nitrate, whereas a white chalky precipitate is formed in case of lead nitrate.
  3. Heat each of the compound with cone, hydrochloric acid. In case of manganese dioxide a greenish yellow gas (chlorine) is evolved. In case of copper(II) oxide, no gas evolved and a bluish green solution is formed.

Question 4.
State one observation : A zinc granule is added to copper sulphate solution.
Answer:
The blue colour of copper sulphate solution fades to form a colourless solution.

Question 5.
Give balanced equation for the reaction :
Silver nitrate solution and Sodium chloride solution.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 11

2013

Question 1.
Give a chemical test to distinguish between :

  1. NaCl soln. and NaNO3 soln.
  2. HCl gas and H2S gas.
  3. Calcium nitrate soln. and zinc nitrate soln.
  4. Carbon dioxide gas and sulphur dioxide gas.

Answer:

  1. Add silver nitrate solution to sodium chloride solution and sodium nitrate solution. In case of sodium chloride, a curdy white ppt. is formed. In case of sodium nitrate solution the reaction mixture remains colourless.
  2. Moist lead acetate paper turns black in case of hydrogen sulphide gas, but does not change its colour in case of hydrogen chloride gas.
  3. To each of the solution add first sodium hydroxide solution in small amount and then in excess. In case of calcium y nitrate a fine white precipitate is formed, which does not dissolve in excess of sodium hydroxide. In case of zinc nitrate a gelatin like white precipitate is formed which dissolves in excess of sodium hydroxide.
  4. To each of the gas add few drops of acidified potassium dichromate solution. In case of carbon dioxide no change takes place. In case of sulphur dioxide, potassium dichromate solution turns blue.

Question 2.
From A : CO ; B : CO2 ; C : NO2 ; D : SO3 – State which will not produce an acid on reaction with water.
Answer:
A : CO

2014

Question 1.
Distinguish between : Sodium nitrate and sodium sulphite (using dilute sulphuric acid)
Answer:
Sodium nitrate will not react with dilute sulphuric acid.
Sodium sulphite reacts with dil. sulphuric acid to liberate foul smelling hydrogen sulphide gas.

Question 2.
State your observation : When moist starch iodide paper is introduced into chlorine gas.
Answer:
The starch iodide paper turns blue due to the liberation of free iodine.

Question 3.
The flame test with a salt P gave a brick red flame. What is the cation in P?
Answer:
Cation in P is Ca2+ (calcium ion).

Question 4.
Gas Q turns moist lead acetate paper silvery black. Identify the gas Q.
pH of liquid R is 10. What kind of substance is R?
Answer:
The gas Q is H2S (Hydrogen sulphide).
The substance R is a alkaline.

2015

Question 1.
Select the gas that has a charateristic rotten egg smell. [ammonia, ethane, hydrogen chloride, hydrogen sulphide, ethyne]
Answer:
Hydrogen sulphide

Question 2.
State one relevant observation : When hydrogen sulphide gas is passed through lead acetate solution.
Answer:
When hydrogen sulphide gas is passed through lead acetate solution, it forms a black precipitate of lead sulphide.

Question 3.
Identify the anion present in each of the following compounds: A, B, C :

  1. Salt ‘A’ reacts with cone. H2SO4 producing gas which fumes in moist air and gives dense fumes with ammonia.
  2. Salt ‘B’ reacts with dil. H2SO4 producing a gas which turns lime water milky but has no effect on acidified potassium dichromate solution.
  3. When barium chloride solution is added to salt solution. ‘C’ a white precipitate insoluble in dilute hydrochloric acid is obtained.

Answer:

  1. Chloride ion (Cl)
  2. Carbonate ion (CO32) or bicarbonate ion (HCO3-1)
  3. Sulphate ion (SO42-)

Question 4.
Identify the cation present in each of the following compounds — W, X, Y, Z :

  1. To solution ‘W’, ammonium hydroxide is added in minimum quantity first and then in excess. A dirty white precipitate is formed which dissolves in excess to form a clear solution.
  2. To solution ‘X’ ammonium hydroxide is added in minimum quantity first and then in excess. A pale blue precipitate is formed which dissolves in excess to form a clear inky blue solution.
  3. To solution ‘Y’ a small amount of sodium hydroxide is added slowly and then in excess. A white insoluble precipitate is formed.
  4. To a salt ‘Z’ calcium hydroxide solution is added and then the mixture is heated. A pungent smelling gas turning moist red litmus paper blue is obtained.

Answer:

  1. Zn2+ (Zinc Ion)
  2. Cu2+ (Copper (II) Ion)
  3. Ca2+ (Calcium Ion)
  4. NH4+ (Ammonium Ion)

2016

Question 1.
Identify the cations in each of the following cases :

  1. NaOH solution when added to the solution ‘A’ gives a reddish brown precipitate.
  2. NH4OH solution when added to the solution ‘B’ gives white ppt which does not dissolve in excess.
  3. NaOH solution when added to the solution ‘C’ gives white ppt which is insoluble in excess.

Answer:

  1. Ferric (Fe3+) ion
  2. Plumbous (Pb2+) ion
  3. Calcium (Ca2+) ion

2017

Question 1.
Choose the correct answer from the options – A chloride which forms a precipitate that is soluble in excess of ammonium hydroxide, is :

A. Calcium chloride
B. Ferrous chloride
C. Ferric chloride
D. Copper chloride.

Answer:

D. Copper chloride.

Question 2.
Identify the substance underlined – Cation that does not form a precipitate with ammonium hydroxide but forms one with sodium hydroxide.
Answer:
Magnesium ions.

Question 3.
Identify the salts P and Q from the observations given below :

  1. On performing the flame test salt P produces a lilac coloured flame and its solution gives a white precipitate with silver nitrate solution, which is soluble in ammonium hydroxide solution.
  2. When dilute HCl is added to a salt Q, a brisk effervescence is produced and the gas turns lime water milky. When NH4OH solution is added to the above mixture [after adding dilute HCl], it produces a white precipitate which is soluble in excess NH4OH solution.

Answer:

(i) The salt P is potassium chloride.

  • Reason : K+ ions give lilac colour to flame and Cl ions react with silver nitrate to form silver chloride precipitate which is soluble in excess of ammonium hydroxide.

(ii) The salt Q is zinc carbonate.

  • Reason : CO3-2 ions are responsible for the liberation of carbon dioxide with HCl. The salt formed is zinc chloride which forms white precipitate with ammonium hydroxide. This precipitate is soluble in excess of ammonium hydroxide.

Additional Questions

Question 1.
The following materials are provided – solutions of cobalt chloride, ammonia, potassium permanganate, lime water, starch-iodide, sodium hydroxide, lead acetate, potassium iodide. Also provided are litmus and filter papers, glowing splinters and glass rods. Using the above how would you distinguish between :

(a) a neutral, acidic and a basic gas
(b) oxygen and hydrogen gas
(c) carbon dioxide and sulphur dioxide gas
(d) chlorine and hydrogen chloride gas
(e) hydrogen sulphide and nitrogen dioxide gas
(f) ammonia and carbon dioxide gas
(g) zinc carbonate and potassium nitrate
(h) hydrated copper sulphate and anhydrous copper sulphate
(i) ammonium sulphate and sodium sulphate.

Answer:

(a) Neutral gas does not effect litmus paper. Acidic gas turns blue litmus paper red and basic gas turns red litmus blue.
(b) Oxygen is obtained by heating KMNO4 whereas hydrogen gas is obtained with the action of Zn and dil H2SO4.
(c) No effect of CO2 on KMnO4 or K2Cr2O7 whereas SO2 turns K2Cr2O7 from orange to green.
(d) Chlorine decolourises the colouring matter whereas HCl does not.
(e) H2S gas turns KMnO4 from pink to colourless and NO2 liberates violet vapours with KI.
(f) NH3 turns red litmus blue and S02 turns blue litmus red.
(g) Lime water turns zinc carbonate milky and no effect on potassium nitrate.
(h) Hydrated copper sulphate anhydrous copper sulphate. Take some dry CuSO4 on filter paper. It will be white in colour.
anhydrous copper sulphate white in colour. CuSO4 (white powder)
Now keep it in air for some time, it will absorb water vapous from atmosphers, its colour will change to blue.
(i) Ammonium sulphate and sodium sulphate : When ammonium sulphate is heated with NaOH, gas ammonia is producedwhich turns red litmus blue. But sodium sulphate has no reaction with NaOH.

Question 2.
Give a chemical test to distinguish between the following:

  1. Sodium carbonate and sodium sulphate
  2. Potassium chloride and potassium nitrate
  3. Copper carbonate and copper sulphite
  4. Lead chloride and lead sulphide
  5. Iron (II) sulphate and iron (III) sulphate
  6. Calcium sulphate and zinc sulphate
  7. Lead nitrate and zinc nitrate
  8. Copper sulphate and calcium sulphate
  9. Manganese dioxide and copper (II) oxide
  10. dil. HCl, dil. HNO3, dil. H2SO4.

[explain the procedure for the preparation of the solutions for the above tests wherever required]
Answer:
(i) Sodium carbonate and sodium sulphate : Add BaCl2 solution to Na2CO3,a white precipitates produced which are soluble in dil. HCl.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 18
(ii) Add cone. H2SO4 to potassium chloride and heat.
colourless gas produced which gives dense-white fumes when a glass rod dipped in ammonia is brought near it.
To the salt solution potassium nitrate, add cone. H2SO4 and copper turnings and heat.
Reddish brown fumes evolved which gives violvet vapours and turns potassium iodide paper brown.
(iii) On adding dilute H2SO4 to copper carbonate acid heating, a colorless odour less gas is evolved which turns lime water milky and has no effect on KMnO4 or K2Cr2O7 solution.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 18.1
CO2 gas turns lime water milky.
Now add dil. H2SO4 to copper sulphite and heat.
A colourless gas with suffocating odour evolved which turns lime water milky and changes then pink colour of acidified KMnO4 to colourless and orange colour of acidified K2Cr2O7 to clear green.
(iv) Added MnO2 and cone. H2SO4 to the salt Lead Chloride and heat it.
A greenish coloured yellow coloured gas with punget odour is evolved which turns moist starch iodide paper blue black.
Now add dil. H2SO4 to Lead sulphide and heat.
Colourless gas with rotten egg smell is evolved. The gas evolved turns moist lead acetate paper silvery black.
(v) Iron (II) sulphate is when reacted with small amount of NaOH solution, dirty green precipitates of Iron (II) hydroxide are produced which are insoluble in excess of NaOH.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 19
Iron (III) sulphate is when reacted with small amount of NaOH solution, reddish brown precipitates of Fe(OH)3 are formed which are insoluble in excess of NaOH.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 19.1
(vi) Add sol. of NaOH to calcium sulphate solution, milky white ppts. of calcium hydroxide are formed which are soluble in excess of NaOH.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 19.2
Now add NaOH to zinc sulphate a white ppts of Zn(OH)2 are produced which are soluble in excess of NaOH.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 19.3
Also when sol. of NH4OH is add to CaSO4, no ppts. are formed i. e., no reaction whereas when a sol. of NH4OH is add to ZnSO4, gelatinous white ppts. are formed which are soluble is excess of NH4OH.
(vii) When a sol. of NH4OH in small amount is added to lead nitrate sol. a chalky white precipitates of lead hydroxide are produced. The precipitats so formed are unsoluble in excess of NH4OH.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 20
Now add NH4(OH) in small amount to zinc nitrate, a Gelatinous white precipitates of Zn(OH)2 are formed which are soluble in excess of NH4OH.
(viii) When a sol. of NaOH in small amount is added to copper sulphate, pale blue precipitates of copper (II) hydroxide are formed which are insoluble in excess of NaOH.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 20.1
Now add sol. of NaOH to calcium sulphate sol. milky white ppts. of Ca(OH)2 are formed which are insoluble in excess of NaOH.
(ix) Test : (i) Add cone. HCl to the sol. and heat, (ii) Filter the sol. after reaction.
Manganese dioxide (Black) : Greenish coloured gas is evolved MnO2 + 4HCl → MnCl2 + 2H2O + ClFilterate is brownish in colour.
Copper (II) oxide : There is no evolution of chlorine gas. CuO → 2HCl + CuCl2 + H2O Filterate is blueish colour.
(x) Dil. HCl is prepared by dissolving HCl gas in water. It ionises to H3O+ and CT ions. HCl gas is prepared by explosive mixture of equal volumes of hydrogen gas and chlorine gas.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 21

Question 3.
Identify the cation (positive ion) and anion (negative ion) in the following substances, A, B and C. Also identify P, Q, R, S, T, U, V, W.
(a) Substance ‘A’is water soluble and gives a curdy white precipitate ‘P’ with silver nitrate solution. ‘P’ is soluble in ammonium hydroxide but insoluble in dil. HNO3. Substance ‘A’ reacts with ammonium hydroxide solution to give a white precipitate ‘Q’ soluble in excess of NH4OH.
(b) A solution of substance ‘B’ is added to barium chloride solution. A white ppt. ‘R’ is formed, insoluble in dil. HCl or HNO3. A dirty green ppt. ‘S’ is formed on addition of ammonium hydroxide to a solution of ‘B’ and the precipitate is insoluble in excess of ammonium hydroxide.
(c) Substance ‘C’ is a coloured, crystalline salt which on heating decomposes leaving a black residue ‘T’. On addition of copper turnings and cone. H2SO2 to ‘C’ a coloured acidic gas ‘U’ is evolved on heating. A solution of ‘C’ is added to NaOH soln. until in excess. A pale blue ppt. ‘P’ is obtained insoluble in excess of NaOH. A solution of ‘C’ then added to NH4 soln. in excess to gives an inky blue solution ‘V’. A solution of ‘C’ is warmed and hydrogen sulphide gas is passed through it. A black ppt. ‘W’ appears.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 22
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 23

Unit Test Paper — Chemistry Practicals

Q.1. Match the ‘cations’ A to F and the solubility of ppt. G or H with the correct colours from ‘X’ and ‘Y’.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 23.1

Answer:
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 23.2
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 24

Q.2. Select the correct ‘anion’ of a salt from the anions given, which matches with description 1 to 5.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 24.1

  1. The salt soln. reacts with AgNO3 soln. to give a white ppt. insoluble in dil. HNO3.
  2. The salt soln. reacts with Ba(NO3)2 soln. to give a white ppt. insoluble in dil. HNO3.
  3. The salt soln. reacts with Ba(NO3)2 soln. to give a white ppt. soluble in dil. HNO3 but insoluble in dil. H2SO4.
  4. The salt reacts with dil. H2SO4 on heating evolving a gas which turns KMnO4 soln. pink to colourless.
  5. The salt reacts with cone. H2SO4 on heating evolving a coloured gas which turns potassium iodide paper brown.

Answer:

  1. Cl ion(D)
  2. SO42-– ion(C)
  3. C032- ion (A)
  4. S2- ion (E)
  5. NO31- ion(B)

Q.3. Give balanced equations for the conversions A and B.
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 24.2
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 25
Answer:
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 25.1

Q.4. Complete the table given below :
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 25.2

Answer:
New Simplified Chemistry Class 10 ICSE Solutions - Practical Chemistry 26

Q.5. Select the correct word from the words in bracket.

  1. The solution which on heating with CaCO3 evolves CO2 gas. [cone. H2SO4 / dil.H2SO4 / dil. HCl]
  2. The solution which can be used to distinguish an ammonium salt from a sodium salt. [CuCl2 soln. / NH4OH / dil. H2SO4 / AgNO3 soln.]
  3. The pH of blood is around 7.4, of saliva is 6.5 and of acid rain is around 4.5. The solution which is slightly alkaline of the three, [saliva / acid rain / blood]
  4. Decomposition of [NaCl / NaHCO3 / NaNO3] by dil. H2SO4, forms an unstable acid.
  5. A metal which reacts with an alkali to liberate hydrogen, [iron / copper / aluminium]

Answer:

  1. dil. HCl
  2. CuCl2soln.
  3. blood
  4. NaHCO3
  5. aluminium.

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New Simplified Chemistry Class 10 ICSE Solutions – Sulphuric Acid

New Simplified Chemistry Class 10 ICSE Solutions – Study Of Compounds – Sulphuric Acid

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QUESTIONS
2000

Question 1.
What do you see when concentrated sulphuric acid is added to copper sulphate 5-water.
Answer:
The colour of blue crystal of CuSO4.5H2O changes to white amorphous as the compound loses its water of crystallisation.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 1
Question 2.
Name one catalyst used industrially which speeds up the conversion of SO2 to SO3 in the production of sulphuric acid in the laboratory or industrially. Write the equation for the conversion of sulphur dioxide to sulphur trioxide. Why does this reaction supply energy. What is the name of the compound formed between SO4 and sulphuric acid.
Answer:
V2O5; It is exothermic reaction ; oleum.

2001

Question 1.
Write equations for:

  1. H2SO4 – producing H2,
  2. Between Pb(NO3)2 and dil. H2SO4.

Answer:

  1.  Zn + H2SO4(dil.) → ZnSO4 + H2
  2. Pb (NO3)2 + H2SO4 (dil.) → PbSO4 ↓+ 2HNO3

Question 2.
Explain how a reagent chosen from: ammonium hydroxide, barium chloride, sodium chloride, sodium hydroxide, sulphuric acid and nitric acid enables to distinguish between the two acids mentioned there in.
Answer:
Barium chloride can be used to distinguish between sulphuric acid and nitric acid. Out of these two acids only sulphuric acid gives a white precipitate with barium chloride solution.

2002

Question 1.
State the substance/s reacted with dilute or concentrated sulphuric acid to form the following gases: 

  1. Hydrogen
  2. Carbon dioxide.

State whether the acid used in each case is dilute or concentrated.
Answer:
(1) Zinc (or any other reactive metal) reacts with dil. H2SO4 to give hydrogen.
Zn (s) + H2SO4 (aq.) → ZnSO4 (aq.) + H2(g)
(2) Sodium carbonate reacts with dil. H2S04 to give C02
Na2C03(s) + H2SO4(aq.) → Na2SO4(aq.) + H2O(l) + CO2(g)
The above reaction can also be carried out with NaHCO3 (sodium bicarbonate) or KHCO3 (potassium bicarbonate)

Question 2.
Write the equations for the laboratory preparation of:

  1.  Sodium sulphate (Na2SO4) using dil. H2SO4,
  2. Lead sulphate (PbSO4) using dil. H2SO4.

Answer:

  1. 2NaOH + H2SO4 (dil) → Na2SO4 + 2H2O
  2. Pb(NO3)2 + H2SO4 (dil) → PbSO4 + 2HNO3

2003

Question 1.
State the name of the process by which H2SO4 is manufactured. Name the catalyst used.
Answer:
By Contact process — vanadium pentoxide (V2O5)

Question 2.
“Concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is………. (less volatile / stronger) in comparison to these two acids.”
Answer:
Less volatile

Question 3.
Write the equations for the laboratory preparation of the following salts using sulphuric acid:

  1. Copper sulphate from copper
  2. Lead sulphate from lead nitrate

Answer:

  1. Cu + 2H2SO4 (dil) → CuSO4 + SO2 + H2O
  2. Pb (NO3)2 + H2SO4  → PbSO4 + 2HNO3

2004

Question 1.
Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide.
Answer:
Platinum or Vanadium pentoxide.

Question 2.
In the Contact process, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two- step procedure is used. Write the equations for the two steps involved.
Answer:
The equations for the two steps involved are:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 2

2005

Question 1.
Write balanced equations for the following:

  1. Potassium hydrogen carbonate and dilute sulphuric acid.
  2. Sodium nitrate and concentrated sulphuric acid.

Answer:

  1. 2KHCO. + H2SO4 (dil) → K2SO4 + 2H2O + 2CO2
  2.  2NaN03 + H2S04 (cone) → Na2S04 + 2HNO3

Question 2.
Choose the property of sulphuric acid (A, B, C or D), which is relevant to each of the preparations
(1) to (2) : A: dil. acid (typical acid properties), B: Non-volatile acid, C: Oxidizing agent, D: Dehydrating agent. Preparation of
(1) HCl
(2) ethene from ethanol
(3) copper sulphate from copper oxide.
Answer:

  1. Non volatile acid (B)
  2. Dehydrating agent (D)
  3. dil. acid (A)

2006

Question 1.
Name the process used for the large scale manufacture of sulphuric acid.
Answer:
Contact process.

Question 2.
Which property of sulphuric acid accounts for its use as a dehydrating agent.
Answer:
Sulphuric acid removes water of crystallization.

Question 3.
H2SO4 is an oxidizing agent and a non volatile acid. Write an equation for each property.
Answer:

  1.  Sulphuric acid as an Oxidising agent —
    C + 2H2 SO4 → CO2 + 2SO2 + 2H2O
  2. Sulphuric acid as an Non-volatile acid —
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 3

Question 4.
Select the correct compound from the list — Ammonia, Copper oxide, Copper sulphate, Hydrogen chloride, Hydrogen sulphide, Lead bromide — This compound smells of rotten eggs.
Answer:
Hydrogen sulphide.

2007

Question 1.
Write balanced equation for the following reactions:

  1. Lead sulphate from lead nitrate solution and dilute sulphuric acid.
  2. Copper sulphate from copper and cone, sulphuric acid.

Answer:

  1. Pb(NO3)2 + H2SO4 (dil) → PbSO4 + 2HNO3
  2. Cu + 2H2SO4→ CuSO4 + SO2 + 2H2O
    (Cone)

Question 2.
Properties of H2SO4 are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v). A : Acid B: Dehydrating agent C: Nonvolatile acid D: Oxidizing agent

  1. C12H12O11+ nH2SO4 → 12C + 11H20 + nH2SO4,
  2. S + 2H2SO4 → 3SO2 + 2H2O,
  3. NaCl + H2SO4 → NaHSO4 + HCl,
  4. CuO + H2SO4 → CuSO4 + H2O
  5. Na2CO3 + H2SO4 Na2SO4 + H2O + CO2
    (Some properties may be repeated)

Answer:
(1) B  (2)  D  (3)  C  (4)  A  (5)  A

Question 3.
Dilute hydrochloric acid and dilute sulphuric acid are both colourless solutions. How will the addition of barium chloride solution to each help to distinguish between the two.
Answer:
Out of dilute hydrochloric acid and dilute sulphuric acid, Dilute hydrochloric acid will give a white ppt. of barium sulphate(BaSO4) with barium chloride solution.
H2S04 (dil.) + BaCl2 (aq.) → BaSO4 (s) + 2HCl
HCl(aq.) + BaCl2 (aq.) → No reaction

Question 4.
From HCl, HNOs, H2SO4, state which has the highest boiling point and which has the lowest.
Answer:
H2SO4 [358°C] has highest boiling point. HCl [-85°C] has lowest boiling point.

2008

Question 1.
Dilute sulphuric acid will produce a white precipitate when added to a solution of:
A.Copper nitrate
B. Zinc nitrate
C. Lead nitrate 
D. Sodium nitrate

Question 2.
Identify the following substances :Liquid E can be dehydrated to product ethene.
Answer:
C2H5OH (Ethanol)

Question 3.
Copy and complete the following table relating to an important industrial processes and its final. Output refers to the product of the process not the intermediate steps.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 4
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 5

Question 4.
Making use only of substances given: dil. sulphuric acid, Sodium carbonate, Zinc, Sodium sulphite, Lead, Calcium carbonate: Give eqautions for the reactions by which you could obtain:
(1) hydrogen
(2) sulphur dioxide
(3) carbon dioxide
(4) zinc carbonate (2 steps)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 6
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 7
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 8

New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 9

Question 5.
What property of cone. H2SO4 is used in the action when sugar turns black in its presence.
Answer:
Cone. Sulphuric acid is a dehydrating agent.

Question 6.
Write the equations for:
(1) dil. H2SO4 and barium chloride.
(2) dil. H2SO4 and sodium sulphide.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 10

Question 7.
Which property of cone. H2SO4 allows it to be used in the preparation of HCl and HNO3
Answer:
Non volatile acid.

2009

Question 1.
Name the gas evolved (formula is not acceptable). – The gas that can be oxidised to sulphur.
Answer:
Hydrogen sulphide (H2S)

2010

Question 1.
Give the equation for:

  1. Heat on sulphur with cone. H2SO4.
  2.  Reaction of – sugar with cone. H2SO4.

Answer:

  1. Reaction of sulphur with cone. H2SO4
    S + 2H2SO4 (cone.) → 3SO2 + 2H2O
  2.  Reaction of sugar with cone. H2SO4
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 11

Question 2.
Give a balanced equation for the conversion of zinc oxide to zinc sulphate.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 12

Question 3.
Select from A, B, C –
A: Sodium hydroxide solution.
B: A weak acid.
C: Dilute sulphuric acid.
The solution which liberates sulphur dioxide gas, from sodium sulphite.

2011

Question 1.
State your observation when – Sugar crystals are added to cone, sulphuric acid.
Answer:
A lot of effervescence takes place in the test tube. The test tube gets very hot. So in the end sugar crystals change in the black residue.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 13

Question 2.
Choose the correct answer from the choices – The gas evolved when dil. sulphuric acid reacts with iron sulphide.
(a) Hydrogen sulphide
(b) Sulphur dioxide
(c) Sulphur trioxide
(d) Vapour of sulphuric acid.

Question 3.
Give a balanced equation for – Dilute sulphuric acid is poured over sodium sulphite
Answer:
Na2SO3 + H2SO4 (dil.) → Na2SO4 + H2O+ SO2

Question 4.
With the help of balanced equations, outline the manufacture of sulphuric acid by the contact process.
Answer:
Contact process: Sulphur or Pyrite Burner
S + O2 → SO2
Contact Tower
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 14
Absorption Tower
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 15
Dilution Tank
H2S2O7 + H2O → 2 H2SO4

Question 5.
State the property of sulphuric acid shown by the reaction of cone, sulphuric acid when heated with
(a) Potassium nitrate
(b) Carbon?
Answer:
(a) It behaves as a non volatile acid and helps in the production of a volatile acid.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 16

(b) It behaves as an oxidising agent and oxidises carbon to carbon dioxide
C + 2H2 SO4→ CO2 ↑+ 2H2O + 2SO2

2012

Question 1.
Name – The gas produced on reaction of dilute sulphuric acid with a metallic sulphide.
Answer:
Hydrogen sulphide (H2S)

Question 2.
Some properties of sulphuric acid are listed below. Choose the role played by sulphuric acid – A, B, C, or D which is responsible for the reactions (i) to (v). Some role/s may be repeated.
A. Dilute acid
B. Dehydrating agent
C. Non-volatile acid
D. Oxidising agent
(1) New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 17
(2) S+H2SO4 (conc) →3SO2 + 2H2O
(3) New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 18
(4) MgO + H2SO4 → MgSO4 + H20
(5) Zn + 2H2SO4 (cone.) →  ZnSO4 + SO2 + 2H2O
Answer:

  1. B: Dehydrating agent
  2. D: Oxidising agent
  3. C: Non-volatile acid
  4. A: Dilute acid
  5. D: Oxidising agent

Question 3.
Give balanced equation for the reaction : Zinc sulphide and dilute sulphuric acid.
Answer:
ZnS + H2 SO4 (dil.) → ZnSO4 + H2S

2013

Question 1.
State one appropriate observation for : Cone. H2SO4 is added to a crystal of hydrated copper sulphate.
Answer:
The blue coloured hydrated copper sulphate crystals disinte­grate with a hissing sound, giving off steam and leaving behind white residue.

Question 2.
In the given equation S + 2H2SO → 3S02 + 2H2O: Identify the role played by cone. H2S04 i.e.
(A) Non-volatile acid
(B) Oxidising agent
(C) Dehydrating agent
(D) None of the above.

Question 3.
Give a balanced equation for : Dehydration of concen­trated sulphuric acid with sugar crystals.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 19

Question 4.
Identify the substance underlined: A dilute mineral acid which forms a white precipitate when treated with barium chloride solution.
Answer:.
Dilute sulphuric acid.

2014

Question 1.
Write balanced equations for the following: Action of concentrated sulphuric acid on carbon.
Answer:
Action of concentrated sulphuric acid on carbon.
C + 2H2SO → CO2 + 2H20 + 2SO2

Question 2.
Distinguish between the following pairs of compounds using the test given within brackets:Dilute sulphuric acid and dilute hydrochloric acid (using barium chloride solution)
Answer:
Out of dilute H2SO4 and dilute HCl, only dilute H2SO4 gives white ppt. of BaSO4 with barium chloride solution
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 20
BaCl2 + HCl → No ppt. formation

Question 3.
State – Any two conditions for the conversion of sulphur dioxide to sulphur trioxide.
Answer:

  1. Two condition for the conversion of SO2 to SO3
  2. The mixture of SO2 gas and O2 gas must be pure and dry and in the ratio of 2 : 1 by volume.
  3. The mixture should be passed over platinised asbestos or vanadium pentaoxide maintained
    at 450 ° C.

Question 4.
Give one equation each to show the following properties  of sulphuric acid:
(1) Dehydrating property.
(2) Acidic nature.
(3) As a non-volatile acid.
Answer:

  1. Dehydrating property.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 21
  2. Acidic nature.
    CuO + H2S04 → CuSO + H2O
  3. As a non-volatile acid.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 22

2015

Question 1.
Identify the acid in each case:

  1. The acid which is used in the preparation of a non­volatile acid.
  2. The acid which produces sugar charcoal from sugar.
  3. The acid on mixing with lead nitrate soln. produces a white ppt. which is insoluble even on heating.

Answer:

  1. Nitric acid (cone.)
  2. Cone, sulphuric acid
  3. Dilute hydrochloric acid

Question 2.
Give equations for the action of sulphuric acid on —
(a) Potassium hydrogen carbonate.
(b) Sulphar
Answer:
(a) Action of sulphuric acid on potassium hydrogen carbonate
2KHCO3 + H2SO4 → K2SO4 + 2H,O + 2CO

(b) Action of sulphuric acid on sulphur
S + 2H2SO4 (cone.) → 3SO2, + 2H2O

Question 3.
In the manufacture of sulphuric acid by the Contact process, give the equations for the conversion of sulphur trioxide to sulphuric acid.
Answer:
In the contact process for the manufacture of sulphuric acid, the equations for the conversion of sulphur trioxide to sulphuric acid are
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 23
H2S2O7 + H2O → 2H2SO4

2016

Question 1.
Write balanced chemical equations for: Action of dilute Sulphuric acid on Sodium Sulphite.
Answer:
Na2SO3 + H2SO4(dil) → Na2SO4 + H2O + SO2

Question 2.
State your observations when:

  1. Barium chloride soln. is mixed with sodium sulphate soln.
  2. Concentrated sulphuric acid is added to sugar crystals.

Answer:
(1)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 24

When sodium sulphate is mixed with barium chloride. White coloured precipitates of Barium sulphate are formed.
(2)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 25

When cone, sulphuric acid is added to sugar crystals black spongy mass (sugar charcoal) is formed.

Question 3.
A, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.
A: Typical acid property
B: Non-volatile acid
C: Oxidizing agent
D: Dehydrating agent
Choose the property (A, B, C or D) depending on which is relevant to each of the following:

  1. Preparation of hydrogen chloride gas.
  2. Preparation of copper sulphate from copper oxide.
  3. Action of cone, sulphuric acid on sulphur.

Answer:

  1. Preparation of Hydrogen chloride gas.
    B: Non-volatile acid
  2. Preparation of Copper sulphate from copper oxide.
    A: Typical acid property
  3. Action of cone. Sulphuric acid on Sulphur.
    C: Oxidizing agent

2017

Question 1.
Write the balanced chemical equation for – Action of concentrated sulphuric acid on sulphur.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 26

Question 2.
State one relevant observation for – Action of cone, sulphuric acid on hydrated copper sulphate.
Answer:
Blue coloured copper sulphate crystals crumble with a hissing sound and change to white powdery mass.

Question 3.
State – How will you distinguish between dilute hydrochloric acid and dilute sulphuric acid using lead nitrate solution.
Answer:
Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution. Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.

Question 4.
Write balanced chemical equations to show –

  • The oxidizing action of cone, sulphuric acid on carbon.
  • The behaviour of H2SO4 as an acid when it reacts with magnesium.
  • The dehydrating property of cone, sulphuric acid with sugar.
  • The conversion of SO3 to sulphuric acid in the Contact process.

Answer:

  1. C + 2H2SO4 (cone.) → CO2 + 2SO2 + 2H2O
  2. Mg + H2SO4 (dil.) → MgSO4 + H2 (g)
  3. C12H22O11 + 11 H2S04(conc.) → 12C + 11H2S04.H2O + ΔH
  4. (a) SO3+ H2SO4 (cone.) → H2S2O7 (oleum)
    (b) H2S2O7 + H2O → 2H2SO4 (cone.)

ADDITIONAL QUESTIONS

Question 1.
State why sulphuric acid was called – ‘oil of vitriol’.
Answer:
Sulphuric acid was initially called ‘oil of vitriol ’.It was initially prepared by – distilling green vitriol [FeSO4.7H2O] and hence the name – ‘oil of vitriol’.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 27
Question 2.
State how you would convert
(1) sulphur
(2) chlorine
(3) sulphur dioxide – to sulphur acid.
Answer:

  1. S + 6HNO. [cone.] → 6NO2 + 2H2O + H2SO4
  2. Cl2 + SO2 + 2H2O → 2HCl + H2SO4
  3. 3SO2 + 2HNO3 + 2H2O → 3H2SO4 + 2NO

Question 3.
State the purpose of the ‘Contact process’.
Answer:
When sulphur is burnt in air, it bums with a pale blue flame forming sulphur dioxide and traces of sulphur trioxide.
S + O2→ SO2 [2S + 3O2 → 2SO4 (traces)]
Burning of sulphur or iron pyrites in oxygen is preferred to purified air since heat energy is wasted in heating the unreactive nitrogen component of the air.

Question 4.
In the Contact process

  1. State how you would convert (a) sulphur (b) iron pyrites to sulphur dioxide in the first step of the Contact process.
  2. State the conditions i.e. catalyst, promoter, temperature and pressure in the catalytic oxidation of sulphur dioxide to sulphur trioxide in the Contact tower. Give a balanced equation for the same.
  3. State why the above catalytic oxidation {reaction supplies energy.
  4. Give a reason why – vanadium pentoxide is preferred to platinum during the catalytic oxidation of sulphur dioxide.
  5. Give a reason why the catalyst mass is heated electrically – only initially.
  6. State why sulphur trioxide vapours are absorbed in concentrated sulphuric acid and not in water to obtain sulphuric acid.

Answer:

  1. S + O2 → 5- SO2
    4FeS2 + 11O2  → 5- 2Fe2O3 + 8SO2
  2. Catalytic oxidation of sulphur dioxide to sulphur trioxide.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 28
    [Above equation for the catalysed reaction is exothermic- hence supplies energy.]
    Catalyst: Vanadium pentoxide [V2O5] or platinum [Pt],
    Temperature: 450-500°C Pressure : 1 to 2 atmospheres
    Conversion ratio: 98% of sulphur dioxide converted to sulphur trioxide.
  3. The catalytic oxidation of SO2 to SO3 is an – exothermic reaction. Thus this reaction supplies energy in the form of heat.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 29
  4. Vanadium pentoxide is preferred – to platinized asbestos as a catalyst since it is comparatively – cheaper and less easily poisoned or susceptible to impurities.
  5. The catalyst – mass is – only initially heated electrically, since the catalytic oxidation of sulphur dioxide is an exothermic reaction and the heat produced maintains the temperature at 450 – 500°C.
  6. Even though sulphur trioxide is an acid anhydride of sulphuric acid – it is not directly absorbed in water to give sulphuric acid.
    The reaction is highly exothermic resulting in production of – a dense fog of sulphuric acid particles which do not condense easily.
    Hence sulphur trioxide vapours are – dissolved in cone, sulphuric acid to give oleum which on dilution with – the requisite amount of soft water in the dilution tank gives – sulphuric acid of the desired concentration [about 98%].

Question 5.
Give a reason why concentrated sulphuric acid is kept in air tight bottles.
Answer:
Concentrated sulphuric acid has a great affinity for water and as such it is a hygroscopic liquid. Being Hygroscopic, it absorbs moisture from the atmosphere and hence cone, sulphuric acid is kept in air tight bottles.

Question 6.
State the basic steps with reasons, involved in diluting a beaker of cone. H2SO4.
Answer:
For dilution of cone. H2SO4, the cone, acid is always added to water and water is never added to the cone, acid even though heat is evolved in both cases.
Reason: If water is added to cone. H2SO4 the heat produced is sufficient to spontaneously vaporise a part of the few drops of water added. This is because the amount of water is very small and bioling point of water is much lower than cone. H2SO4, which is in bulk. Due to this sudden vaporisation of water cone, acid tend to spurt out and cause serious injuries.
On the other hand, if cone. H2SO4 is added to water, the heat produced in this case can only raise the temperature of water slightly because water is an bulk. Thus, in this case spurting of the cone, acid is avoided.

Question 7.
Give reasons why dilute sulphuric acid:
(1) behaves as an acid when dilute.
(2) is dibasic in nature.
Answer:

  1. Acidic properties of sulphuric acid are due to the presence of – hydronium ions [H3O+ ] formed when H2SOdissociates in aq. solution.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 30
  2. Sulphuric acid dissociates in – aq. solution giving 2H+ ions per molecule – of  the acids .Hence its – basically is two.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 31
    [Basicity is the number of H+ ions formed by dissociation of one molecule of the acid in its aq. soln.]

Question 8.
Convert dil. H.SO, to –

  1. Hydrogen
  2. Carbon dioxide
  3. Sulphur dioxide
  4. Hydrogen sulphide
  5. An acid salt
  6.  A normal salt.

Answer:

  1. Dil H2SO4 to hydrogen by the action of any active metal (say zinc) on dil. H2SO4
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 32
  2. Dil. H2SO4t0 carbon dioxide by the action of any carbonate or bicarbonate on dil. H2SO4
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 33
  3. Dil. H2S04to sulphur dioxide by the action of any sulphide on dil. H2SO4
    Na2SO3 + H2SO4 → Na2SO4 + H2O + SO2
  4. H2SOto hydrogen sulphide by the action of any sulphide on dil. H2SO4
    FeS + H2SO4 (dil.) → FeSO4 + H2S↑
  5. H2S04to an acid by the action of insufficient strong base with excess of dil. H2SO4
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 34
  6. H2SO4to a normal salt by the action of sufficient (or excess of) strong base (NaOH) with excess of dil. FI2SO4
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 35

Question 9.
Give equations for formation of two different acids from cone. H2SO4. State the property of sulphuric acid involved in the above formation.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 36
Property of Sulphuric acid involved in the formation of these acids: Cone. H2SO4 is a non-volatile acid.

Question 10.
Give equations for oxidation of cone. H2SO4 giving the oxidised products –

  1. Carbon dioxide
  2. Sulphur dioxide
  3. Phosphoric acid
  4. Copper (II) sulphate
  5. Iodine
  6. Sulphur respectively.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 37

Question 11.
Give a reason why concentrated and not dil H2SO4 – behave as an oxidising and dehydrating agent.
Answer:

  1. Cone. H2SO4 be aves as an oxidising agent because cone.H2SO4 whes h. ated decomposes to give nascent oxygen which acts as a strong oxidising agent.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 38
    On the other hand, dil H2SO4 on heating does not decompose to give nascent oxygen and as such cannot behave as an oxidising agent.
  2. Cone. H2SO4 behaves as a dehydrating agent because cone. H2SO4 act as a great affinity for water and hydration of cone. H2SO4 is an exothermic reaction.On the other hand, dil H2SO4 has no affinity for water and hence cannot act as a hydrating agent.

Question 12.
Give the equation for the reaction cone, sulphuric acid with –

  1.  glucose
  2. sucrose
  3. cellulose
  4. an organic acid containing one carbon atom and two hydrogen atoms
  5. an organic acid containing two carbon and two hydrogen atoms
  6. an alcohol
  7. hydrated copper (II) sulphate.

Answer:

  1. Reaction of cone. H2SO4 with glucose, CtH12O6
    Dehydration.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 39
  2.  Reaction of cone. H2SO4 with sucrose, C12H12O12 –
    Dehydration.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 40
  3. Reaction of cone. H2SO4 with cellulose, (C6H10O5)n –
    Dehydration.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 41
  4. Reaction of cone. H2SO4 with an organic acid containing one carbon atom and two hydrogen atoms, HCOOH (formic acid or methanoic acid) –
    Dehydration.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 42
  5. Reaction of cone. H2SO4 with an organic acid containing two carbon atom and two hydrogen atoms, [COOH]2 (oxalic acid or erhanedi ic acid) –
    Dehydration.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 43
  6. Reaction of cone. H2SO4 with an alcohol (other than methanol) – say ethyl alcohol or ethanol, c2H5OH
    Dehydration.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 44
  7. Reaction of cone. H2SO4 with hydrated copper (II) sulphate, CuS04. 5H2O (blue vitriol) – Dehydration.New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 45


Question 13.

State the observation seen when cone. H2SO4 is added to –
(1) sucrose
(2 ) hydrated copper (II) sulphate.
Answer:

  1. Cone. H2SO4 dehydrates sucrose to carbon, called sugar charcoal. This is m the form of a black spongy charged mass of carbon.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 46
  2. Cone. H2S04 dehydrates blue crystals of copper (II) sulphate pentahydrate (blue vitriol) to copper sulphite, which is in the form of a white powder.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 47

Question 14.
State how addition of –

  1. copper
  2. NaCl to hot cone. H2SO4 serves as a test for the latter.

Answer:

  1. Copper turnings when heated with cone. H2SO4 gives a colourless suffocating gas with a smell of burning sulphur(SO2). The gas turns orange coloured acidified potassium dichromate solution green. This can be used as a test for cone. H2SO4.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 48
  2. Common salt (NaCl) when heated with cone. H2SO4 gives a colourless gas pungent smell only which (HCl) gives white fumes with NH3. This can be used as a test for cone. H2SO4.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 49

Question 15.
Give two tests for dilute sulphuric acid with balanced equations. State why
(1) BaCl2
(2)  Pb(NO3)2 are used for the above tests.
Answer:

  1. Barrium chloride solution on treating with dilute sulphuric acid forms white ppt. of barium sulphate, which is insoluble in all acids
    BaCl2 + H2SO4 (dil) → 2HCl + BaS04
  2. Lead nitrate on treating with dilute sulphuric acid forms white ppt of lead sulphate, which is insoluble in all acids.
    Pb (NO3)2 + H2SO4 (dil) → 2HNO3 + PbSO4

Question 16.
Give a test to distinguish dilute sulphuric acid from dilute HCl and dilute HNO3.
Answer:
Test to distinguish dil H2SO4 from dil HCl and HNO3- BaCl2 soln. when added to dil. H2S04 gives a white ppt. of BaSO4, but with dil. HCl and dil. HNO3, no white ppt. is produced – since BaCl2 and Ba(NO3)2 are soluble in dil. H2SO4.

Question 17.
State three different chemical compounds other than acids manufactured industrially from sulphuric acid.
Answer:

  1. Barium sulphate
  2. Lead sulphate
  3. Sodium sulphate

UNIT TEST PAPER 7D — H2SO4

Question 1.
Select the correct answer from the choice in brackets.

  1. The oxidised product obtained when sulphur reacts with cone. H2S04. (H2S/SO2/H2SO3).
    Ans. SO2
  1. The dehydrated product obtained when cane sugar reacts with cone. H2S04. (CO / C / CO2)
    Ans. C
  1. The type of salt formed when excess of caustic soda reacts with sulphuric acid, (acid salt / normal salt)
    Ans. Normal salt
  1. The reduced product obtained when hydrogen sulphide reacts with cone. H2S04. (SO, / S / H.O)
    Ans. SO2
  1. The salt which reacts with dil. H2SO4 acid to give an insoluble ppt. (Cu (NO3)2 / Zn (NOs)2 / Pb (NO3)2
    Ans. Pb(NO3)2

Question 2
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 50

  1. Give a balanced equation for the conversion ‘A’.
    Ans. 4FeS2 + 11O2→ 2Fe2O3 + 8SO2
  2. The gaseous mixture of the product of conversion ‘A’ and air contains dust particles as an impurity. Name another impurity in the same mixture.
    Ans: Arsenious oxide (As2O3)
  3. Is the conversion ‘B’ an exothermic or an endothermic reaction. Would lowering the temperature favour or retard the forward reaction.
    Ans. Conversion of SO2 into SO3 is an exothermic reaction. As such lowering of temperature will favour the forward reaction i.e. Formation of SO3.
  4. If the product of conversion ‘B’ is an acid anhydride of H2S04, the anhydride of conversion ‘A’ is………..
    Ans. Acidic.
  5. State why water is added for the conversion ‘D’ and not for the conversion ‘C’
    Ans. SO3 is not directly dissolved in water to give H2SO4. This is because the dissolution of SO3 in water is highly exothermic resulting in production of dense fog of sulphuric acid particles which do not condense easily.

Question 3.
Give balanced equations for the following reactions using sulphuric acid.

  1. Formation of a black mark on a piece of wood on addition of cone. H2SO4 to it.
    Ans. (C6H10O5)n + H2S04 (cone.) → 6(C)n + 5(H2O)n
  2. Oxidation of a foul smelling acidic gas, heavier than air and fairly soluble in H2O by cone. H2SO4.
    Ans. H2S + H,SO4 (cone.) → S + 2H2O + SO2
  3. Formation of an acid salt from sulphuric acid and (a) an alkali (b) a sodium salt.
    Ans.
    (a) Formation of an acid salt from sulphuric acid and an alkali (KOH)
    KOH + H2SO4 (dil.)→ KHSO4 + 2H2O
    (b) Formation of an acid salt from sulphuric acid and a sodium salt (NaCl)
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 51
  4. Formation of a hydrocarbon from an organic compound
    Ans:
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 52
  5. Formation of sulphur dioxide using a metal below hydrogen in the activity series.
    Ans: Cu + 2H2SO4 (cone.) → CuSO4 + 2H2O +SO2

Question 4.
Match the conversions in column ‘X’ using sulphuric acid, with the type of chemical property of sulphuric acid A to E it represents in column ‘Y’
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 53
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 54

Question 5.
Select the correct substance from the substances A to J which react with the sulphuric acid to give the product 1 to 10. [State whether the acid used in each case is dilute or concentrated].
A : Iron
B : Sodium carbonate
C : Sodium chloride
D : Formic acid
E : Sodium nitrate
F : Sodium sulphite
G : Ethyl alcohol
H : Sodium sulphide
I : Sodium hydroxide (excess)
J : Hydrogen sulphide

  1. Product – ulphur dioxide
  2. Product- Sulphur
  3. Product-Hydrogen
  4. Product-Hydrochloric acid
  5. Product-Sodium sulphate
  6. Product-Carbon dioxide
  7. Product-Carbon monoxide
  8. Product-Nitric acid
  9. Product—Hydrogen sulphide
  10. Prorfwcf-Ethene

Answer:

  1. F: Sodium sulphite
  2. H: Sodium sulphide
  3. A: Iron
  4. C: Sodium chloride
  5. I: Sodium hydroxide
  6. E: Sodium Carbonate
  7. D: Formic acid
  8. E: Sodium nitrate
  9. H: Sodium sulphide
  10. G: Ithyl alcohol

Question 6.
Give reasons for the following:

  1. Sulphuric acid forms two types of salts with an alkali.
    Ans. Sulphuric acid forms two types of . disc, viz., sulphates and bisulphates (or hydrogen sulphates) with alkales because it is a dibasic :acid, i.e. one molecule of H2SO4 on dissociation gives  two H+ ions.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 55
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 56
  2. Cone, sulphuric acid is used as a laLoraioiy reagent in the preparation of iodine from hydrogen iodide.
    Ans. Cone. H2SO4. oxidises HI to iodine.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 57
  3. Barium chloride solution can be used to distinguish between dil. H2SO4 and dil HNO3.
    Ans. BaCl2 soln. when added to dil. H2SO4 gives a white ppt. of BaSO4, but with dil. HNO3, n: white ppt. is produced since BaCl2 and Ba(NO3), are soluble in dil. H2SO4.
  4. The gaseous product obtained differs when zinc reacts with dilute and with cone. H2SO4
    Ans. The metal reacts differently with dilute acid and concentrated acid. That is the character of metal. With dilute sulphuric acid, zinc gives hydrogen.
    Zn + H2SO4 (dil.) → ZnSO4 + H2
    The same metals will react differently with concentrated sulphuric acid to give sulphur dioxide gas.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 58
  5. Ethanol can be converted to ethene using cone, sulphuric acid.
    Ans. Ethanol cart be converted into ethene by heating it with cone. H2SO4 because cone. H2SO4 is a strong dehydrating agent.
    New Simplified Chemistry Class 10 ICSE Solutions Chapter 7D Study Of Compounds - Sulphuric Acid 59

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New Simplified Chemistry Class 10 ICSE Solutions – Stoichiometry

New Simplified Chemistry Class 10 ICSE Solutions – Mole Concept and Stoichiometry : Percentage Composition – Empirical & Molecular Formula Calculations Based on Chemical

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Additional Problems

Q.1. Percentage Composition

Question 1.
Calculate the percentage by weight of :
(a) C in carbon dioxide
(b) Na in sodium carbonate
(c) Al in aluminium nitride.
[C = 12, O = 16, H = 1, Na = 23, Al = 27, N = 14]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 1
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 2

Question 2.
Calculate the percentage of iron in K3Fe(CN)6. [K = 39,Fe = 56, C = 12, N = 14]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 3

Question 3.
Calculate which of the following – calcium nitrate or ammonium sulphate has a higher % of nitrogen.[Ca = 40, S = 32, N = 14, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 4
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 5

Question 4.
Calculate the percentage of pure aluminium in 10kg. of aluminium oxide [Al2O3] of 90% purity. [Al = 27, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 6

Question 5.
State which of the following are better fertilizers —

  1. Potassium phosphate [K3PO4] or potassium nitrate [KNO3]
  2. Urea [NH2CONH2] or ammonium phosphate [(NH4)3PO4]
    [K = 39, P = 31, O = 16, N = 14, H = 1]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 7
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 8

Question 6.
Calculate the percentage of carbon in a 55% pure sample of carbon carbonate. [Ca = 40, C = 12, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 9

Question 7.
Calculate the percentage of water of crystallisation in hydrated copper sulphate [CuSO4.5H2O].
[Cu = 63.5, S = 32, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 10

Question 8.
Hydrated calcium sulphate [CaSO4xH2O] contains 21% of water of crystallisation. Calculate the number of molecules of water of crystallisation Le. ‘X’ in the hydrated compound.
[Ca = 40, S = 32, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 11
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 12

Q.2. Empirical And Molecular Formula

Question 1.
A compound gave the following data : C = 57.82%, O = 38.58% and the rest hydrogen. Its vapour density is 83. Find its empirical and molecular formula. [C = 12, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 13

Question 2.
Four g of a metallic chloride contains 1.89 g of the metal ‘X’. Calculate the empirical formula of the metallic chloride. [At. wt. of ‘X’ = 64, Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 14

Question 3.
Calculate the molecular formula of a compound whose empirical formula is CH2O and vapour density is 30.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 15

Question 4.
A compound has the following percentage composition. Al = 0.2675g.; P = 0.3505g.; O = 0.682g. If the molecular weight of the compound is 122 and its original weight on analysis gave the above results 1.30 g. Calculate the molecular formula of the compounds
[Al = 27, P = 31, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 16

Question 5.
Two organic compounds ‘X’ and ‘Y’ containing carbon and hydrogen only have vapour densities 13 and 39 respectively. State the molecular formula of ‘X’ and ‘Y’ [C = 12, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 17

Question 6.
A compound has the following % composition. Zn = 22.65%; S = 11.15%; O = 61.32% and H = 4.88%. Its relative molecular mass is 287 g. Calculate its molecular formula assuming that all the hydrogen in the compound is present in combination with oxygen as water of crystallization.
[Zn = 65, S — 32, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 18

Question 7.
A hydrocarbon contains 82.8% of carbon. Find its molecular formula if its vapour density is 29 [H = 1, C = 12]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 19

Question 8.
An organic compound on analysis gave H = 6.48% and O = 51.42%. Determine its empirical formula if the compound contains 12 atoms of carbon.
[C = 12, H = 1, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 20
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 21

Question 9.
A hydrated salt contains Cu = 25.50%, S = 12.90%, O = 25.60% and the remaining % is water of crystallization. Calculate the empirical formula of the salt.
[Cu = 64, S = 32, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 22

Question 10.
A gaseous hydrocarbon weights 0.70 g. and contains 0.60 g. of carbon. Find the molecular formula of the compound if its molecular weight is 70.
[C = 12, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 23

Question 11.
A salt has the following % composition Al = 10.50%, K = 15.1%, S = 24.8% and the remaining oxygen. Calculate the empirical formula of the salt.
[Al = 27, K = 39, S = 32, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 24
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 25

Q.3. Chemical Equations

Question 1.
What mass of silver chloride will be obtained by adding an excess of hydrochloric acid to a solution of 0.34 g of silver nitrate.
[Cl = 35.5, Ag = 108, N = 14, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 26

Question 2.
What volume of oxygen at s.t.p. will be obtained by the action of heat on 20 g
KClO3 [K = 39, Cl = 35.5, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 27

Question 3.
From the equation : 3Cu + 8HNO3 → 3Cu(NO3)2 + 4H2O + 2NO. Calculate

  1. the mass of copper needed to react with 63 g of nitric acid
  2. the volume of nitric oxide collected at the same time. [Cu = 64, H = 1, O = 16, N = 14]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 28

Question 4.
Zinc blende [ZnS] is roated in air. Calculate :

(a) The number of moles of sulphur dioxide liberated by 776 g of ZnS and
(b) The weight of ZnS required to produce 22.4 lits. of SO2 at s.t.p. [S = 32, O = 16, Zn = 65]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 29

Question 5.
Ammonia reacts with sulphuric acid to give the fertilizer ammonium sulphate. Calculate the volume of ammonia [at s.t.p.] used to form 59 g of ammonium sulphate.
[N = 14, H = 1, S = 32, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 30

Question 6.
Heat on lead nitrate gives yellow lead [II] oxide, nitrogen dioxide and oxygen. Calculate the total volume of NO2 and O2 produced on heating 8.5 of lead nitrate.
[Pb = 207, N = 14, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 31

Question 7.
Calculate the amount of KClO3 which on thermal decomposition gives ‘X’ vol. of O2, which is the volume required for combustion of 24 g. of carbon.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 32
[K = 39, Cl = 35.5, O = 16, C = 12].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 33

Question 8.
Calculate the weight of ammonia gas.
(a) Required for reacting with sulphuric acid to give 78g. of fertilizer ammonium sulphate.
(b) Obtained when 32.6g. of ammonium chloride reacts with calcium hydroxide during the laboratory preparation of ammonia.
[2NH4Cl + Ca(OH)2 → CaCl2 + 2H2O + 2NH3] [N = 14, H = 1, O = 16, S = 32, Cl = 35.5].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 34
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 35

Question 9.
Sodium carbonate reacts with dil. H2SO4 to give the respective salt, water and carbon dioxide. Calculate the mass of pure salt formed when 300 g. of Na2CO3 of 80% purity reacts with dil. H2SO4.
[Na = 23, C = 12, O = 16, H = 1, S = 32].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 36

Question 10.
Sulphur burns in oxygen to give sulphur dioxide. If 16 g. of sulphur burns in ‘x’ cc. of oxygen, calculate the amount of potassium nitrate which must be heated to produce V cc. of oxygen. [S = 32, K = 39, N = 14, O = 16].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 37

Question 11.
Sample of impure magnesium is reacted with dilute sulphuric acid to give the respective salt and hydrogen. If 1 g. of the impure sample gave 298.6 cc. of hydrogen at s.t.p. Calculate the % purity of the sample. [Mg = 24, H = 1].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 38

Questions

A. Lussac’S Law – Problems based on them

2006

Question 1.
560 ml. of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows : 2CO + O2 → 2CO2. Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 39
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 40

2009

Question 1.
200 cm3 of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.
2C2H2(g) + 5O2 (g) → 4CO2 (g) + 2H2O (g)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 41

2010

Question 1.
10 litre of this mixture is burnt, And the total volume of carbon dioxide gas added to the atmosphere. Combustion reaction can be represented as :
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 42
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 43
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 44

2012

Question 1.
67.2 litres of H2 combines with 44.8 litres of N2 to form NH3 :
N2(g) + 3H2(g) → 2NH3(g). Calculate the vol. of NH3 produced. What is the substance, if any, that remains in the resultant mixture ?
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 45

2013

Question 1.
What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?
2C4H10 + 13O2 → 8CO2 + 10H2O
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 46

2014

Question 1.
What volume of ethyne gas at s.t.p. is required to produce 8.4 dm3 of carbon dioxide at s.t.p.?
2C2H2 + 5O2 → 4CO2 + 2H2O [H = 1, C = 12, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 47

2015

Question 1.
If 6 litres of hydrogen and 4 litres of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 48
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 49

2016

Question 1.
The equations 4NH3 + 5O2 → 4NO + 6H2O, represents the catalytic oxidation of ammonia. If 100 cm3 of ammonia is used calculate the volume of oxygen required to oxidise the ammonia completely.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 50

2017

Question 1.
Propane burns in air according to the following equation :
C3H8 + 5O2 → 3CO2 + 4H2O. What volume of propane is consumed on using 1000 cm3 of air, considering only 20% of air contains oxygen.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 51
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 52

B. Mole Concept – Avogadro’S Number – Problems based on them

2004

Question 1.
A flask contains 3.2 g of SO2. Calculate :

  1. The moles of SO2 and the no. of molecules of SO2 present in the flask.
  2. The volume occupied by 3.2 g. of SO2 at s.t.p. (S = 32, O = 16)

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 53

Question 2.
2KMnO4 → K2MnO4 + MnO2 + O2 Given that the molecular mass KMn04 is 158, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g. of potassium permanganate. (Molar volume at room temperature is 24 litres.)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 54

2005

1. The volumes of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temp, and press.

(i) Which sample contains the maximum number of molecules. If the temp, and pressure of gas A are kept constant, then what will happen to the volume of A when the no. of molecules is doubled.
Answer:
The sample D and volume of A will get Doubled.

(ii) If this ratio of gas vols. refers to reactants and products of reaction – gas law observed is ….
Answer:
Gay Lussac’s law of combining volumes.

(iii) If the volume of ‘A’ is 5.6 dm3 at s.t.p., calculate the no. of molecules in the actual vol. of ‘D’ at s.t.p. (Avog no. is 6 × 1023). Using your answer, state the mass of ‘D’ if the gas is “N2O” (N = 14, O = 16). [6 x 1023, 44g.]
Answer:
Vol. of D will be 4 × 5.6 = 22.4 lit. and 22.4 lit. of D contain molecules = 6 × 1023 (AV. number)
Mass of N2 O (i.e. D) = (14 × 2 + 16 × 1) = 44g

2006

Question 1.
Calculate the no. of moles and the no. of molecules present in 1.4 g of ethylene gas (C2H4). What is the vol. occupied by the same amount of C2H4. State the vapour density of C2H4.
(Avog. No. = 6 × 1023 ; C = 12, H = 1]
(0.05 moles ; 3 × 1022 molecules ; 1.12 lit. ; 14)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 56
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 57

2008

1. The equation for the burning of octane is : 2C8H18 + 25O2 → 16CO2 + 18H2O

(i) How many moles of carbon dioxide are produced when one mole of octane burns ?
Answer:
2 moles of octane produce 16 moles of C02
1 mole of octane will produce 16/2 moles of C02 = 8 moles of carbon dioxide.

(ii) What volume, at s.t.p., is occupied by the number of moles determined in (1) (i) ?
Answer:
1 moles occupy a volume of 22.4 litres.
8 mole will occupy a volume of = 22.4 × 8 = 179.2 litres

(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane ?
Answer:
From equation, we know that 2moles of octane produces 16 moles of CO2.
Mass of CO2 produced = 16 × 44 = 704

2009

Question 1.
Define the term – Mole. A gas cylinder contains 24 × 1024 molecules of nitrogen gas. If Avogadro’s number is 6 × 1023 and the relative atomic mass of nitrogen is 14, calculate :

  1. Mass of nitrogen gas in the cylinder.
  2. Volume of nitrogen at STP in dm3

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 58

Question 2.
Gas ‘X’ occupies a volume of 100 cm3 at S.T.P and weighs 0.5 g. find its relative molecular mass.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 59

2010

Question 1.
Dilute hydrochloric acid (HCl) is reacted with 4.5 moles of calcium carbonate. Give the equation for the said reaction. Calculate :

  1. The mass of 4.5 moles of CaCO3.
  2. The volume of CO2 liberated at stp.
  3. The mass of CaCl2 formed ?
  4. The number of moles of the acid HCl used in the reaction (relative molecular mass of CaCO3 is 100 and of CaCl2 is 111].

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 60

2011

Question 1.
Calculate the mass of :

  1. 1022 atoms of sulphur.
  2. 0.1 mole of carbon dioxide.
    [S = 32, C = 12 and O = 16 and Avogadro’s Number 6 × 1023]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 61

Question 2.
Calculate the volume of 320 g of SO2 at stp. [S = 32 and O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 62
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 63

2012

Question 1.
The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 64

2013

Question 1.
The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP ?
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 65

Question 2.
Calculate the volume occupied by 0.01 mole of CO2 at STP.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 66

2014

Question 1.
State Avogadro’s Law. A cylinder contains 68g of ammonia gas at s.t.p.

  1. What is the volume occupied by this gas?
  2. How many moles and how many molecules ammonia are present in the cylinder? [N = 14, H = 1]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 67

2015

Question 1.
From A, B, C, D, which weighs the least —
A : 2 g. atoms of Nitrogen
B : 1 mole of Silver
C : 22.4 litres of oxygen gas at 1 atmospheric pressure and 273K
D : 6.02 × 1o23 atoms of carbon.
[Ag = 108, N=14, 0=16, C=12]
Answer:
D : 6.02 × 1023 atoms of carbon.

Question 2.
Calculate the mass of Calcium that will contain the same number of atoms as are present in 3.2 gm of sulphur. [S = 32, Ca = 40]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 68

2016

Question 1.
Select the correct answer from A, B, C and D : The ratio between the number of molecules in 2g of hydrogen and 32g of oxygen is: k

(A) 1 : 2
(B) 1 : 0.01
(C) 1:1
(D) 0.01 : 1
[H = 1, O = 16]

Answer:

(C) 1 : 1

Question 2.
A gas of mass 32 gms has a volume of 20 litres at S.T.P. Calculate the gram mol. weight of the gas.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 69

Question 3.
A gas cylinder contains 12 x 1024 molecules of oxygen gas. Calculate :

  1. the mass of 02 present in the cylinder.
  2. the volume of O2 at S.T.P. present in the cylinder.[0 = 16] Avog. no. is 6 × 1023 [640g. ; 448 l]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 70
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 71

2017

Question 1.
Calculate the number of gram atoms in 4.6 grams of sodium [Na = 23]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 72

Question 2.
The mass of 11.2 litres of a certain gas at s.t.p. is 24 g. Find the gram molecular mass of the gas.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 73

C. Mole Concept – Avogadro’S Law – problmes based on them

2001

Question 1.
The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. If 20 litres of nitrogen contain ‘X’ no. of molecules state the no. of molecules in 10 litres of chlorine, 20 litres of ammonia and 5 litres of sulphur dioxide, (x/2, x, x/4)
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 74
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 75

2002

Question 1.
Samples of gases O2, N2, CO and CO2 under the same conditions of temperature and pressure contain the same number of molecules represented by X. The molecules of oxygen occupy V litres and have a mass of 8 g. Under the same conditions of temperature and pressure, what is the volume occupied by :

  1. X molecules of N2.
  2. 3X molecules of CO.
  3. What is the mass of CO2 in grams.
  4. In answering the above questions, whose law has been used.
    (C = 12, N = 14, O = 16)

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 76
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 77

2005

Question 1.
Define the term atomic weight :
Answer:
Atomic weight : is the number of times one atom of an element is heavier than 1/2 the mass of an atom of carbon (C12)

2008

Question 1.
The gas law which relates the volume of a gas to the number of molecules of the gas is :

A Avogadro’s Law
B Gay-Lussac’s Law
C Boyle’s Law
D Charle’s Law

Answer:

A Avogadro’s Law

2009

Question 1.
Correct the following – Equal masses of all gases under identical conditions contain the same number of molecules.
Answer:
Equal volumes of all gases under identical conditions contain the same number of molecules.

2013

Question 1.
A vessel contains X number of molecules of hydrogen gas at a certain temperature & pressure. Under the same conditions of temperature & pressure, how many molecules of nitrogen gas would be present in the same vessel.
Answer:
According to Avogadro’s law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.
Hence, number of molecules of N2 = Number of molecules of H2 = X

2017

Question 1.
A gas cylinder can hold 1 kg. of H2 at room temp. & press.:

  1. Find the number of moles of hydrogen present
  2. What weight of CO2 can the cylinder hold under similar conditions to temp. & press
  3. If the number of molecules of hydrogen in the cylinder is X, calculate the number of CO2 molecules in the cylinder under the same conditions of temp. & press
  4. State the law that helped you to arrive at the above result.

Answer:

  1. 2 g of hydrogen gas = 1 mole.
    ∴ 1000 g of hydrogen gas 1000/2 = 500 moles.
  2. 1 mole of carbon dioxide = 44 g
    ∴ 500 moles of carbon dioxide = 44 x 500 = 22000 g = 22 kg Weight of carbon dioxide in cylinder = 22 kg
  3. Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
    ∴ Molecules in the cylinder of carbon dioxide = X.
  4. Avogadro’s law.

D. Vapour Density & Molecular Weight – Problems based on them

1996

Question 1.
Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87°C and 380 mm Hg pressure. (1 litre of hydrogen at s.t.p. weighs 0.09 g)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 78
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 79

2001

Question 1.
Mention the term defined by the following : The mass of a given volume of gas compared to the mass of an equal volume of hydrogen.
Answer:
Vapour Density.

2004

Question 1.
Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in a mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g. Calculate the relative molecular mass of oxygen.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 80
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 81
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 82

2009

Question 1.
A gas cylinder of capacity of 20 dm3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2g, hence the relative molecular mass of the gas is :

(A) 5
(B) 10
(C) 15
(D) 20

Answer:

(B) 10

2012

Question 1.
The vapour density of carbon dioxide [C = 12, O = 16] is

(A) 32
(B) 16
(C) 44
(D) 22

Answer:

(D) 22

2014

Question 1.
Give one word or phrase for : The ratio of the mass of a certain volume of gas to the mass of an equal volume of hydrogen under the same conditions of temperature and pressure.
Answer:
Vapour density.

E. Percentage Composition – Problems based on them

1996

Question 1.
Find the total percentage of oxygen in magnesium nitrate crystals : Mg(NO3)2. 6H2O
(H = 1, N = 14, O = 16, Mg = 24)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 83

1997

Question 1.
What is the mass of nitrogen in 1000 kg of urea [CO(NH2)2].[C = 12] (Answer to the nearest kg.)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 84

1998

Question 1.
Calculate the % of boron (B) in borax Na2 B4 O7. 10H2O. (H = 1, B = 11, O = 16, Na = 23)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 85

1999

Question 1.
If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass in kg. of the fertilizer calcium nitrate would be required to replace the nitrogen in a 10 hectare field. (N = 14 ; O = 16 ; Ca = 40)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 86

2001

Question 1.
Calculate the percentage of phosphorus in the fertilizer superphosphate Ca(H2PO4)2. (correct to 1dp) (H = 1 ; O = 16 ; P = 31 ; Ca = 40)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 87

2002

Question 1.
Calculate the percentage of platinum in ammonium chloroplatinate (NH4)2 PtCl6
(Give your answer correct to the nearest whole number). (N = 14, H = 1, Cl = 35.5, Pt = 195)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 88

2005

Question 1.
Calculate the percentage of nitrogen in aluminium nitride. (Al = 27, N = 14)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 89

2006

Question 1.
Calculate the percentage of sodium in sodium aluminium fluoride (Na3AlF6) correct to the nearest whole number. (F = 19 ; Na = 23 ; A1 = 27)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 90

2007

Question 1.
Determine the percentage of oxygen in ammonium nitrate (O = 16)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 91

2010

Question 1.
If the relative molecular mass of ammonium nitrate is 80, calculate the percentage of nitrogen and oxygen in ammonium nitrate. (N = 14, H = 1, O = 16).
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 92

2012

Question 1.
Find the total percentage of Magnesium in magnesium nitrate crystals, Mg(NO3)2.6H2O/ [Mg = 24; N = 14; O = 16 and H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 93

2017

Question 1.
Calculate the percentage of water of crystallization in CUSO4.5H2O. [H = 1, O = 16, S = 32, Cu = 64]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 94

F. Empirical Formula And Molecular Formula – Problems based on then

2000

Question 1.
Determine the empirical formula of the compound whose composition by mass is : 42% nitrogen 48% oxygen and 9% hydrogen.
(H = 1 ; N = 14 ; O 16).
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 95

2001

Question 1.
A metal M forms a voltaic chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen (i.e. V.D.) is 162.5, find the molecular formula of the chloride. (M = 56, Cl = 35.5)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 96
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 97

2002

Question 1.
The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound (work to two decimal places). (Na = 23, P = 31, O = 16)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 98

2004

Question 1.
An experiment showed that in a lead chloride solution, 6.21 g of lead combined with 4.26 g. of chlorine. What is the empirical formula of this chloride. (Pb = 207 ; Cl = 35.5)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 99
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 100

2006

Question 1.
Determine the empirical formula of a compound containing 47.9% potassium, 5.5% beryllium and 46.6% fluorine by mass.
(At. weight of Be = 9 ; F = 19 ; K = 39) Work to one decimal place.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 101

2007

Question 1.
A compound X consists of 4.8% carbon and 95.2% bromine by mass

  1. Determine the empirical formula of this compound working correct to one decimal place
  2. If the vapour density of the compound is 252, what is the molecular formula of the compound. (C = 12 ; Br = 80) (CBr3, C2Br6).

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 102

2008

Question 1.
What is the empirical formula of octane (C8H18) ?
Answer:
M.F. of octane = C8H18 = (c4H9)2
∴ E.F. of octane = C4H9

Question 2.
A compound has the following percentage composition by mass, Carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. The relative molecular mass of this compound is 168, so what is its molecular formula?
(H = 1 ; C = 12 ; Cl = 35.5)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 103

2009

Question 1.
A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 37.
[N =14, H = 1].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 104
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 105

2011

Question 1.
An organic compound with vapour density 94. It contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula of the organic compound.
[C = 12, H = 1, Br = 80]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 106

2014

Question 1.
A compound having empirical formula X2Y is made of two elements X and Y. Find its molecular formula. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density 25.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 107

2015

Question 1.
If the empirical formula of a compound is CH and it has a vapour density of 13, find the molecular formula of the compound. [C = 12, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 108

2016

Question 1.
A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapours density is 29, find its molecular formula. [C=12,H=1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 109
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 110

2017

Question 1.
A compound of X and Y has the empirical formula XY2. Its vapou density is equal to its empirical formula weight. Determine its Molecular formla.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 111

G Chemical Equations – Problems based on them

2000

Question 1.
Washing soda has the formula Na2CO3.10H2O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 112

Question 2.
Na2SO4 + Pb(N03)2 → PbSO4 + 2NaNO3 When excess » lead nitrate solution was added to a solution of sodium sulphate, 15.15 g of lead sulphate were precipitated. What mass of sodium sulphate was present in the original solution. (H = 1 ; C = 12 ; O = 16 ; Na = 23; S = 32 ; Pb = 207)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 113

2001

Question 1.
From the equation : (NH4)2 Cr2O7 → Cr2O3 + 4H2O + N2 Calculate :

  1. the vol. of nitrogen at STP, evolved when 63g. of ammonium dichromate is heated. (5.6 lits.)
  2. the mass of Cr2O3 formed at the same time. (N = 14, H = 1, Cr = 52, O = 16) (38 g.)

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 114

2003

Question 1.
10g. of a mixture of NaCI and anhydrous Na2SO4 is dissolved in water. An excess of BaCl2 soln. is added and 6.99 g. of BaSO is precipitated according to the equation :
Na2SO4 + BaCl2 → BaSO4 ↓+ 2NaCl. Calculate the percentage of sodium sulphate in the original mixture.
(O = 16 ; Na = 23 ; S = 32; Ba = 137)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 115

2004

Question 1.
The reaction of potassium permanganate with acidified iron (II) sulphate is given below :
2KMnO4 + 10FeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5 Fe2(SO4)3 + 8H2O.
If 15.8 g. of potassium permanganate was used in the reaction, calculate the mass of iron (II) sulphate used in the above reaction.
(K = 39, Mn = 55, Fe = 56, S = 32, O = 16)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 116
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 117

2005

Question 1.
The equation given below relate to the manufacture of sodium carbonate (Mol), wt. of Na2CO3 = 106

  1. NaCl + NH3 + CO2 + H2O → NaHCO3 + NH4Cl
  2. 2NaHCO3 → Na2CO3 + H2O + CO2

Questions (a) and (b) are based on the production of 21.2 g. of sodium carbonate.
(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium carbonate (Molecular weight of NaHCO3 = 84).
(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 118

2006

Question 1.
The relative molecular mass (mol. wt.) of copper oxide is 80. What vol. of HN3 (measured at s.t.p.) is required to completely reduce 120 g of cuO.
(3CuO + 2NH3 → 3Cu + 3H2O + N2].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 119

2007

Question 1.
A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at stp.)
NH4 NO3 → N2O + 2H2O

  1. What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam.
  2. What mass of ammonium nitrate should be heated to produce 8.96 litres of steam (Relative molecular mass of NH4NO3 is 80)

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 120

2008

Question 1.
From the equation :
C + 2H2SO4 → CO2 + 2H2O + 2SO2
Calculate :

  1. The mass of carbon oxidised by 49 g of sulphuric acid (C = 12 ; relative molecular mass of sulphuric acid = 98).
  2. The volume of SO2 measured at s.t.p., liberated at the same time.

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 121

2009

Question 1.
Commercial NaOH weighing 30 g. has some NaCl in it. The mixture on dissolving in water and treatment with excess AgNO3 soln. formed a precipitate weighing 14.3 g. What is the percentage of NaCl in the commercial sample of NaOH.
NaCl + AgNO3 → AgCl + NaNO3
[Relative molecular mass of NaCl = 58 ; AgCl = 143]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 122

2011

Question 1.
Calculate the volume of oxygen required for the complete combustion of 8.8g of propane (C3 H8). (C = 12, O = 16, H = 1, Molar Volume = 22.4 dm3 at stp)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 123

2013

Question 1.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 124
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 125

2015

Question 1.
From the equation :
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 126

Calculate:

  1. the quantity in moles of (NH4)2Cr2O7 if 63gm oft[NH4)2Cr2O7 is heated.
  2. the quantity in moles of nitrogen formed.
  3. the volume in litres or dmof N2 evolved at s.t.p.
  4. the mass in grams of Cr2O3 formed at the same time. [H=l, Cr= 52, N=14]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 127
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 128

2016

Question 1.
How much calcium oxide is formed when 82g. of calcium nitrate is heated. Also find the volume of nitrogen dioxide evolved :
2Ca(NO3)2 → 2CaO + 4NO2 + O2 (Ca = 40, N = 14, O = 16)
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4B Mole Concept and Stoichiometry Percentage Composition - Empirical & Molecular Formula 129

For More Resources

New Simplified Chemistry Class 10 ICSE Solutions – Mole Concept

New Simplified Chemistry Class 10 ICSE Solutions – Mole Concept and Stoichiometry : Gay Lussac’s Law – Avogadro’s Law

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions

Viraf J Dalal Chemistry Class 10 Solutions and Answers

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Practice Questions

Question 1.
What volume of oxygen would he required to burn completely 400 ml of acetylene (C2H2) ? Calculate the volume of CO2 formed.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 1

Question 2.
2500 cc of oxygen was burnt with 600 cc of ethane (C2H6). Calculate the volume of unused oxygen and the volume of carbon dioxide formed, after writing the balanced equation :
Ethane + Oxygen → Carbon dioxide + Water vapour
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 2
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 3

Question 3.
80 cm3 of methane are mixed with 200 cm3 of pure oxygen at similar temperature and pressure. The mixture is then ignited. Calculate the composition of resulting mixture if it is cooled to initial room temperature and pressure.
CH4 + 2O2 → CO2 + 2H2O
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 4

Question 4.
Calculate the volume of HCl gas formed and chlorine gas required when 40 mL of methane reacts completely with chlorine at S.T.P.
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 5

Question 5.
What volume of propane is burnt for every 500 cm3 of air used in the reaction under the same conditions ?
C3H8 + 5O2 → 3CO2 + 4H2O
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 6

Question 6.
450 cm3 of nitrogen monoxide and 200 cm3 of oxygen are mixed together and ignited. Calculate the composition of resulting mixture.
2NO + O2 → 2NO2
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 7

Question 7.
24 cc marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc of which 58 cc were unchanged oxygen. Which law does their experiment supports ? Explain with calculations.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 8
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 9
Avogadro’s Law : “Under the same conditions of temperature and pressure equal volumes of all gases contain die same number of molecules.”
If we assume that 7 litre of oxygen gas contains ‘n’ molecules of the gas then by Avogadro’s Law :

  1. 1 litre of oxygen will contain ‘n’ molecules of hydrogen
  2. 1 litre of nitrogen will contain ‘n’ molecules of nitrogen.
  3. 1 litre of any gas will contain ‘n’ molecules of that gas. e.g.

New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 10

Relative Molecular Mass : “It is the number that represents how many times one molecule of a substance is heavier than one atom of hydrogen whose weight has been taken unity or 1/12 of 6C12.
Relative Molecular Mass : “It is the number that represents how many times one molecule of a substance is heavier than one atom of hydrogen whose weight has been taken unity or 1/12 of 6C12.
Avogadro’s Number : “The number of atoms present in 12 g (gm atomic mass) of 6C12 is called Avogadro’s number.
Na or L = 6.023 × 1023

New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 11
Mole : Mole is the mass of substance containing particles equal to Avogadro’s number i.e. 6.023 × 1023.
Gram Atom : “The relative atomic mass of an element expressed in grams is called gram atom.
Gram Mole : “The relative atomic mass of a substance expressed in grams is called gram mole.
Molar Volume : Volume occupied by one mole of any gas at STP is called molar volume.

Applications Of Avogadro’S Law :

  1. Determines the atomicity of the gas.
  2. Determines the molecular formula of a gas.
  3. Determines the relation between molecular weight and vapour density.
  4. Explains Gay-Lussac’s law.
  5. Determines the relationship between gram molecular weight and gram molecular volume.

1. Determines the atomicity of a gas :
Atomicity : The number of atoms present in one molecule of that element is called atomicity.
Monoatomic : Elements which have one atom in their molecules e.g. Helium, Neon.
Diatomic : Elements which have two atoms in their molecule e.g. Hydrogen, nitrogen, oxygen.
e.g. of Determination of atomicity of a gas :
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 12
A molecules of nitrogen contains two atoms : atomicity – Diatomic

2. Determines the molecular formula of a gas.
Molecular formula : A chemical formula which gives the actual or exact number of atoms of the elements present in one molecule of a compound.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 13

3. Determines the relation between molecular weight and vapour density.
Molecular weight: It is the ratio of the weight of 1 molecule of a substance to the weight of one atom of hydrogen.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 14
Vapour Density : It is the ratio of the mass of a certain volume of gas or vapour of the mass of the same volume of hydrogen.
Mol. wt. = 2× vapour density

4. Explains Gay-Lussac’s Law :
Gay-Lussac’s Law is explained by Avogadro’s Law which states “Under similar conditions of temperature and pressure, equal volumes of different gases have same number of molecules.”
Since substances react in simple ratio of number of molecules, volumes of gaseous reactants and products will also bear a simple ratio to one another. This is what Gay-Lussac’s Law says.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 15

5. Determines relationship between gram molecular mass and gram molecular volume :
Gram molecular mass is the relative molecular mass of a substance expressed in grams. It is also called gram molecule of that element.
Gram molecular volume : The volume occupied by e.g. molecular wt. of a gas at s.t.p.
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 16

Additional Problems

Q.1. Lussac’S Law

Question 1.
Nitrogen reacts with hydrogen to give ammonia. Calculate the volume of the ammonia gas formed when nitrogen reacts with 6 litres of hydrogen. All volumes measured at s.t.p.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 17

Question 2.
2500 cc of oxygen was burnt with 600 cc of ethane [C2H6]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 18

Question 3.
20 ml. each of oxygen and hydrogen and 10 ml. of carbon monoxide are exploded in an enclosure. What will be the volume and composition of the mixture of the gases when they are cooled to room temperature.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 19

Question 4.
224 cm3 of ammonia undergoes catalytic oxidation in presence of Pt to given nitric oxide and water vapour. Calculate the volume of oxygen required for the reaction. All volumes measured at room temperature and pressure.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 20
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 21

Question 5.
Acetylene [C2H2] burns in air forming carbon dioxide and water vapour. Calculate the volume of air required to completely burn 50 cm3 of acetylene. [Assume air contains 20% oxygen].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 22

Question 6.
On igniting a mixture of acetylene [C2H2] and oxygen, 200 cm3 of CO2 is collected at s.t.p. Calculate the volume of acetylene & O2 at s.t.p. in the original mixture.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 23

Question 7.
Ammonia is formed from the reactants nitrogen and hydrogen in presence of a catalyst under suitable conditions. Assuming all volumes are measured in litres at s.t.p. Calculate the volume of ammonia formed if only 10% conversion has taken place.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 24

Question 8.
100 cc. each of water gas and oxygen are ignited and the resultant mixture of gases cooled to room temp. Calculate the composition of the resultant mixture. [Water gas contains CO and H2 in equal ratio]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 25

Q.2. Mole Concept – Avogadro’S Law – Avogadro’S Number
Calculate the following : [all measurement at s.t.p. or as stated in the problem]

Question 1.
The mass of 2.8 litres of C02. [C = 12, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 27
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 28

Question 2.
The volume occupied by 53.5g of Cl2. [Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 29

Question 3.
The number of molecules in 109.5 g of HCl. [H = 1, Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 30

Question 4.
The number of

  1. molecules [S = 32]
  2. atoms in 192 g. of sulphur. [S8]

Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 31

Question 5.
The mass of (Na) sodium which will contain 6.023 × 1023 atoms. [Na = 23]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 32

Question 6.
The no. of atoms of potassium present in 117 g. of K. [K = 39]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 33

Question 7.
The number of moles and molecules in 19.S6 g. of Pb (NO3)2. [Pb = 207, N = 14, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 34

Question 8.
The mass of an atom of lead [Pb = 202].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 35

Question 9.
The number of molecules in 1 1/2 litres of water. [density of water 1.0 g./cm3. — ∴ mass of water = volume × density]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 36
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 37

Question 10.
The gram-atoms in 88.75 g of chlorine [Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 38

Question 11.
The number of hydrogen atoms in 0.25 mole of H2SO4.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 39

Question 12.
The gram molecules in 21 g of nitrogen [N = 14]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 40

Question 13.
The number of atoms in 10 litres of ammonia [N = 14, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 41
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 42

Question 14.
The number of atoms in 60 g of neon [Ne = 20]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 43

Question 15.
The number of moles of ‘X’ atoms in 93 g of ‘X’ [X is phosphorus = 31]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 44

Question 16.
The Volume occupied by 3.5 g of O2 gas at 27°C and 740 mm presure. [O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 45
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 46

Question 17.
The moles of sodium hydroxide contained in 160 g of it. [Na = 23, O = 16, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 47

Question 18.
The weight in g. of 2.5 moles of ethane [C2H6]. [C = 12, H = 1]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 48

Question 19.
The molecular weight of 2.6 g of a gas which occupies 2.24 lits. at 0°C and 760 mm press.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 49

Question 20.
The gram atoms in 46 g of sodium [Na = 23]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 50

Question 21.
The number of moles of KCl03 that will be required to give 6 moles of oxygen.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 51

Question 22.
The weight of the substance of its molecular weight is 70 and in the gaseous form occupies 10 lits. at 27°C and 700 mm pressure.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 52

Question 23.
Has higher number of moles : 5 g. of N2O or 5 g. of NO [N = 14, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 53

Question 24.
Has higher mass : 1 mole of CO2 or 1 mole of CO [C = 12, O = 16]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 54

Question 25.
Has higher no. of atoms : 1 g of O2 or 1 g of Cl2 [O = 16, Cl = 35.5]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 55

Q.3. Vapour Density And Molecular Weight

Question 1.
500 ml. of gas ‘X’ at s.t.p. weighs 0.50 g. Calculate the vapour density and molecular weight of the gas. [1 lit. of H2 at s.t.p. weighs 0.09 g].
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 56

Question 2.
A gas cylinder holds 85 g of a gas ‘X’. The same cylinder when filled with hydrogen holds 8.5 g of hydrogen under the same conditions of temperature and pressure Calculate the molecular weight of ‘X’.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 57

Question 3.
Calculate the relative molecular mass [molecular weight] of 290 ml. of a gas ‘A’ at 17°C and 1520 mm pressure which weighs 2.73 g at s.t.p. [1 litre of hydrogen at s.t.p. weighs 0.09 g.]
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 58
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 59

Question 4.
State the volume occupied by 40 g of a hydrocarbon – CH4 at s.t.p. if its V.D. is 8.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 60

Question 5.
Calculate the atomcity of a gas X [at. no. 35.5] whose vapour density is equal to its relative atomic mass.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 61

Question 6.
Calculate the relative molcular mass and vapour density of methyl alcohol [CH3OH] if 160 g. of the alcohol on vaporization has a volume of 112 litres at s.t.p.
Answer:
New Simplified Chemistry Class 10 ICSE Solutions Chapter 4A Mole Concept and Stoichiometry Gay Lussac’s Law - Avogadro’s Law 62

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