Selina ICSE Solutions for Class 10 Chemistry

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

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Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 4 Analytical Chemistry: Uses Of Ammonium Hydroxide And Sodium Hydroxide. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 4 Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide

Exercise 1

Solution 1.

(a) Ferrous salts : Light green
(b) Ammonium salts : Colourless
(c) Cupric salts : Blue
(d) Calcium salts : Colourless
(e) Aluminium salts : Colourless

Solution 2.

(a) Cu(OH)2
(b) ZnO
(c) NaOH
(d) NH4OH
(e) Na+, Ca2+
(f) Fe2+, Mn2+
(g) Aluminium
(h) Zn(OH)2 and Al(OH)3
(i) PbO
(j) Ammonium ion

Solution 3.

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Solution 4.

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Solution 5.

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Solution 6.

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Solution 7.

(a) ZnCl2
(b) Zn(OH)2

Solution 8.

(a) PbO
(b) ZnO
(c) K2ZnO2

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 5

Solution 10.

When freshly precipitated aluminum hydroxide reacts with caustic soda solution, whitesalt of sodium meta aluminate is obtained.
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 6

Solution 11.
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 7

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 8
With excess of NaOH, white gelatinous ppt. of Zn (OH)2 is soluble. So, these two cannot be distinguished by NaOH alone. However white ppt. of Pb(OH)2 is readily soluble in acetic acid also.
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 9
(i) On addition of NH4OH to calcium salts no precipitation of Ca(OH)2 occurs even with addition of excess of NH4OH because the concentration of OH–ions from ionization of NH4OH is so low that it cannot precipitate the hydroxide of calcium.
Pb(NO3)2+2 NH4OHPb(OH)2+2NH4NO3
Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 10
Solution 13.

Lead carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is insoluble in excess.
Zinc carbonate is dissolved in dilute nitric acid and then ammonium hydroxide is added to it. A white precipitate is formed which is soluble in excess.

Solution 14.

Reagent bottles A and B can identified by using calcium salts such as Ca(NO3)2.

On adding NaOH to Ca (NO3)2, Ca (OH)2 is precipitated as white precipitate which is sparingly soluble in excess of NaOH.
Ca(NO3)2+2NaOH → Ca(OH)2+ 2NaNO3

Whereas, on addition of NH4OH to calcium salts, no precipitation of Ca(OH)2 occurs even with addition of excess of NH4OH because the concentration of OH–ions from the ionization of NH4OH is so low that it cannot precipitate the hydroxide of calcium.
So the reagent bottle which gives white precipitate is NaOH and the other is NH4OH.

Intext Exercise

Solution 1.

(i) Analysis: The determination of chemical components in a given sample is called analysis.
(ii) Qualitative analysis: The analysis which involves the identification of the unknown substances in a given sample is called qualitative analysis.
(iii) Reagent: A reagent is a substance that reacts with another substance.
(iv) Precipitation: It is the process of formation of an insoluble solid when solutions are mixed. The solid thus formed is called precipitate.

Solution 2.

(i) Yellow
(ii) Colourless
(iii) PaleGreen
(iv) Colourless
(v) Colourless

Solution 3.

(i) Fe3+
(ii) Cu2+
(iii) Cu+2
(iv) Mn2+

Solution 4.

(i) Ca(OH)2
(ii) Fe(OH)2 and Cu(OH)2
(iii) Zn(OH)2 and Pb(OH)2

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Analytical Chemistry: Uses of Ammonium Hydroxide And Sodium Hydroxide img 11

Solution 6.

NH4OH and NaOH can be distinguished by using calcium salts.

For example on adding NaOH to Ca(NO3)2, Ca(OH)2 is obtained as white precipitate which is sparingly soluble in excess of NaOH.

Ca(NO3)2 + 2NaOH → Ca(OH)2 + 2NaNO3

On addition of NH4OH to calcium salts, no precipitation of Ca(OH)2 occurs even with the addition of excess of NH4OH.This is because the concentration of OH– ions from the ionization of NH4OH is so low that it cannot precipitate the hydroxide of calcium.

Solution 7.

(i) Fe(OH)2 and Pb(OH)2
(ii) Cu(OH)2 and Zn(OH)2

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Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry

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Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 5 Mole Concept and Stoichiometry

Exercise 5(A)

Solution 1.

(a) Gay-Lussac’s law states that when gases react, they do so in volumes which bear a simple ratio to one another, and to the volume of the gaseous product, provided that all the volumes are measured at the same temperature and pressure.

(b) Avogadro’s law states that equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Solution 2.

a) The number of atoms in a molecule of an element is called its atomicity. Atomicity of Hydrogen is 2, phosphorus is 4 and sulphur is 8.

b) N2means 1 molecule of nitrogen and 2N means two atoms of nitrogen.
N2 can exist independently but 2N cannot exist independently.

Solution 3.

(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
Now volume of hydrogen gas =volume of helium gas
n molecules of hydrogen =n molecules of helium gas
nH2=nHe
1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium
Therefore 2H=He
Therefore atoms in hydrogen is double the atoms of helium.

(b) For a given volume of gas under given temperature and pressure, a change in any one of the variable i.e., pressure or temperature changes the volume.

(c) Inflating a balloon seems violating Boyles law as volume is increasing with increase in pressure. Since the mass of gas is also increasing.

Solution 4.

2H2 + O2 → 2H2O
2 V     1V         2V

From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm3 of Hydrogen reacts with = 200/2= 100 cm3
Hence, the unreacted oxygen is 150 – 100 = 50cm3 of oxygen.

Solution 5.

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Solution 6.

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Solution 7.

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Solution 8.

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Solution 9.

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Solution 10.

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Solution 11.

C3H8 + 5O2 → 3CO2 + 4H2O
1 V 5 V 3 V

From equation, 5 V of O2 required = 1V of propane
so, 100 cm3 of O2 will require = 20 cm3 of propane

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 7

Solution 13.

2CO + O2 → 2CO2
2 V 1 V 2 V

2 V of CO requires = 1V of O2
so, 100 litres of CO requires = 50 litre of O2

Solution 14.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 8

Solution 15.

H2 + Cl2 → 2HCl
1V 1V 2 V

Since 1 V hydrogen requires 1 V of oxygen and 4cm3 of H2 remained behind so the mixture had com”>16 cm3 hydrogen and 16 cm3 chlorine.
Therefore Resulting mixture is H2 =4cm3,HCl=32cm3

Solution 16.

CH4 + 2O2 → CO2 + 2H2O
1 V 2 V 1 V

2C2H2 + 5O2 → 4CO2 + 2H2O
2 V 5 V 4 V

From the equations, we can see that
1V CH4 requires oxygen = 2 V O2
So, 10cm3 CH4 will require =20 cm3 O2
Similarly 2 V C2H2 requires = 5 V O2
So, 10 cm3 C2H2 will require = 25 cm3 O2
Now, 20 V O2 will be present in 100 V air and 25 V O2 will be present in 125 V air ,so the volume of air required is 225cm3

Solution 17.

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Solution 18.

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Solution 19.

This experiment supports Gay lussac’s law of combining volumes.
Since the unchanged or remaining O2 is 58 cc so, used oxygen 106 – 58 = 48cc
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.

CH4 + 2O2 → CO2 + 2H2O
1 V 2 V
24 cc 48 cc
i.e. methane and oxygen react in a 1:2 ratio.

Solution 19.

According to Avogadro’s law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules.
So,
(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.
(b) 1 litre of SO2 contains the least number of molecules since it has the smallest volume.

Solution 20.

GasVolume (in litres)Number of molecules
Chlorine10x/2
Nitrogen20x
Ammonia20X
Sulphur dioxide5x/4

Solution 21.

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Exercise 5(B)

Solution 1.

a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.
b) The value of avogadro’s number is 6.023 × 1023
c) The molar volume of a gas at STP is 22.4 dm3 at STP

Solution 2.

(a) The vapour density is the ratio between the masses of equal volumes of gas and hydrogen under the conditions of standard temperature and pressure.

(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm3.

(c) The relative atomic mass of an element is the number of times one atom of the element is heavier than 1/12 times of the mass of an atom of carbon-12.

(d) The relative molecular mass of an compound is the number that represents how many times one moleculae of the substance is heavier than 1/12 of the mass of an atom of carbon-12.

(e) The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x1023 atoms.

(f) The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element.

(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

Solution 3.

(a) Applications of Avogadro’s Law :

  1. It explains Gay-Lussac’s law.
  2. It determines atomicity of the gases.
  3. It determines the molecular formula of a gas.
  4. It determines the relation between molecular mass and vapour density.
  5. It gives the relationship between gram molecular mass and gram molecular volume.

(b) According to Avogadro’s law under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.

Since substances react in simple ratio by number of molecules, volumes of the gaseous reactants and products will also bear a simple ratio to one another.This what Gay Lussac’s Law says.

H2 + Cl2 → 2HCl
1V 1V 2V (By Gay-Lussacs law)
n molecules n molecules 2n molecules (By Avogadros law)

Solution 4.

(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5 x 6 = 444
(b) KClO3 = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H2O)5 x 18 = 249.5
(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132
(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82
(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5
(g) (2N)28 + (8H)8 + (2Cr)2 x 51.9+ (7O)7 x 16 = 252

Solution 5.

(a) No. of molecules in 73 g HCl = 6.023 x1023 x 73/36.5(mol. mass of HCl)
= 12.04 x 1023

(b) Weight of 0.5 mole of O2 is = 32(mol. Mass of O2) x 0.5=16 g

(c) No. of molecules in 1.8 g H2O = 6.023 x 1023 x 1.8/18
= 6.023 x 1022

(d) No. of moles in 10g of CaCO3 = 10/100(mol. Mass CaCO3)
= 0.1 mole

(e) Weight of 0.2 mole H2 gas = 2(Mol. Mass) x 0.2 = 0.4 g

(f) No. of molecules in 3.2 g of SO2 = 6.023 x 1023 x 3.2/64
= 3.023 x 1022

Solution 6.

Molecular mass of H2O is 18, CO2 is 44, NH3 is 17 and CO is 28
So, the weight of 1 mole of CO2 is more than the other three.

Solution 7.

4g of NH3 having minimum molecular mass contain maximum molecules.

Solution 8.

a) No. of particles in s1 mole = 6.023 x 1023
So, particles in 0.1 mole = 6.023 x 10 23 x 0.1 = 6.023 x 1022

b) 1 mole of H2SO4 contains =2 x 6.023 x 1023
So, 0.1 mole of H2SO4 contains =2 x 6.023 x 1023 x0.1
= 1.2×1023 atoms of hydrogen

c) 111g CaCl2 contains = 6.023 x 1023 molecules
So, 1000 g contains = 5.42 x 1024 molecules

Solution 9.

(a) 1 mole of aluminium has mass = 27 g
So, 0.2 mole of aluminium has mass = 0.2 x 27 = 5.4 g
(b) 0.1 mole of HCl has mass = 0.1 x 36.5 (mass of 1 mole)
= 3.65 g
(c) 0.2 mole of H2O has mass = 0.2 x 18 = 3.6 g
(d) 0.1 mole of CO2 has mass = 0.1 x 44 = 4.4 g

Solution 10.

(a) 5.6 litres of gas at STP has mass = 12 g
So, 22.4 litre (molar volume) has mass =12 x 22.4/5.6
= 48g(molar mass)
(b) 1 mole of SO2 has volume = 22.4 litres
So, 2 moles will have = 22.4 x 2 = 44.8 litre

Solution 11.

(a) 1 mole of CO2 contains O2 = 32g
So, CO2 having 8 gm of O2 has no. of moles = 8/32 = 0.25 moles
(b) 16 g of methane has no. of moles = 1
So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles

Solution 12.

(a) 6.023 x 10 23 atoms of oxygen has mass = 16 g
So, 1 atom has mass = 16/6.023 x 1023 = 2.656 x 10-23 g
(b) 1 atom of Hydrogen has mass = 1/6.023 x 1023 = 1.666 x 10-24
(c) 1 molecule of NH3 has mass = 17/6.023 x1023 = 2.82 x 10-23 g
(d) 1 atom of silver has mass = 108/6.023 x 1023 =1.701 x 10-22
(e) 1 molecule of O2 has mass = 32/6.023 x 1023 = 5.314 x 10-23 g
(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g

Solution 13.

(a) 0.1 mole of CaCO3 has mass =100(molar mass) x 0.1=10 g
(b) 0.1 mole of Na2SO4.10H2O has mass = 322 x 0.1 = 32.2 g
(c) 0.1 mole of CaCl2 has mass = 111 x 0.1 = 11.1g
(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g

Solution 14.

1molecule of Na2CO3.10H2O contains oxygen atoms = 13
So, 6.023 x1023 molecules (1mole) has atoms=13 x 6.023 x 1023
So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 1023 =7.8 x 1023

Solution 15.

3.2 g of S has number of atoms = 6.023 x1023 x 3.2 /32
= 0.6023 x 1023
So, 0.6023 x 1023 atoms of Ca has mass=40 x0.6023×1023/6.023 x 1023
= 4g

Solution 16.

(a) No. of atoms = 52 x 6.023 x1023 = 3.131 x 1025
(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms = 6.023 x1023
So, 52 g will have = 6.023 x 1023 x 52/4 = 7.828 x1024 atoms

Solution 17.

Molecular mass of Na2CO3 = 106 g
106 g has 2 x 6.023 x1023 atoms of Na
So, 5.3g will have = 2 x 6.023 x1023x 5.3/106=6.022 x1022 atoms
Number of atoms of C = 6.023 x1023 x 5.3/106 = 3.01 x 1022 atoms
And atoms of O = 3 x 6.023 x 1023 x 5.3/106= 9.03 x1022 atoms

Solution 18.

(a) 60 g urea has mass of nitrogen(N2) = 28 g
So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg
(b) 64 g has volume = 22.4 litre
So, 320 g will have volume = 22.4 x 320/64=112 litres

Solution 19.

(a) Vapour density of carbon dioxide is 22, it means that 1 molecule of carbon dioxide is 22 heavier than 1 molecule of hydrogen.
(b) Vapour density of Chlorine atom is 35.5.

Solution 20.

22400 cm3 of CO has mass = 28 g
So, 56 cm3 will have mass = 56 x 28/22400 = 0.07 g

Solution 21.

18 g of water has number of molecules = 6.023 x 1023
So, 0.09 g of water will have no. of molecules = 6.023 x 1023 x 0.09/18 = 3.01 x 1021 molecules

Solution 22.

(a) No. of moles in 256 g S8 = 1 mole
So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles

(b) No. of molecules = 0.02 x 6.023 x 1023 = 1.2 x 1022 molecules
No. of atoms in 1 molecule of S = 8
So, no. of atoms in 1.2 x 1022 molecules = 1.2 x 1022 x 8
= 9.635x 1022 molecules

Solution 23.

Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of P4 = 123.88 g
If phosphorus is considered as P4 molecules,
then 1 mole P4 ≡ 123.88 g
Therefore, 100 g of P4 = 0.807 g

Solution 24.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 12

Solution 25.

No. of atoms in 12 g C = 6.023 x1023
So, no. of carbon atoms in 10-12 g = 10-12 x 6.023 x1023/12
= 5.019 x 1010 atoms

Solution 26.

Given:
P= 1140 mm Hg
Density = D = 2.4 g / L
T = 273 0C = 273+273 =   546 K
M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L
Hence we have to find out the volume of the unknown gas at STP.

First apply Charle’s law.
We have to find out the volume of one liter of unknown gas at standard temperature 273 K.

V1= 1 L  T1 = 546 K
V2=?       T2 = 273 K
V1/T1 = V2/ T2
V2 = (V1 x T2)/T1
      = (1 L x 273 K)/546 K
= 0.5 L

We have found out the volume at standard temperature. Now we have to find out the volume at standard pressure.

Apply Boyle’s law.
P 1 = 1140 mm Hg  V1 = 0.5 L
P2 = 760 mm Hg  V2 = ?
P1 x V1 = P2 x V2
V2 = (P1 x V1)/P2
      = (1140 mm Hg x 0.5 L)/760 mm Hg
= 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP
X moles = 0.75 L / 22.4 L
=  0.0335 moles
The original mass is 2.4 g
n = m / M
0.0335 moles    = 2.4 g / M
M = 2.4 g / 0.0335 moles
M= 71.6 g / mole
Hence, the gram molecular mass of the unknown gas is 71.6 g

Solution 27.

1000 g of sugar costs = Rs. 40
So, 342g(molar mass) of sugar will cost=342×40/1000=Rs. 13.68

Solution 28.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 13
Solution 29.

40 g of NaOH contains 6.023 x 1023 molecules
So, 4 g of NaOH contains = 6.02 x1023 x 4/40
= 6.02 x1022 molecules

Solution 30.

The number of molecules in 18 g of ammonia= 6.02 x1023
So, no. of molecules in 4.25 g of ammonia = 6.02 x 1023 x 4.25/18
= 1.5 x 1023

Solution 31.

(a) One mole of chlorine contains 6.023 x 1023 atoms of chlorine.
(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.
(c) Relative atomic mass of an element is the number of times one atom of an element is heavier than 1/12 the mass of an atom of carbon-12.
(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

Exercise 5(C)

Solution 1.

Information conveyed by H2O

  1. That H2O contains 2 volumes of hydrogen and 1 volume of oxygen.
  2. That ratio by weight of hydrogen and oxygen is 1:8.
  3. That molecular weight of H2O is 18g.

Solution 2.

The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.
The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

Solution 3.

(a) CH (b) CH2O (c) CH (d) CH2O

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 14

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 15

Solution 6.

Molecular mass of KClO3 = 122.5 g
% of K = 39 /122.5 = 31.8%
% of Cl = 35.5/122.5 = 28.98%
% of O = 3 x 16/122.5 = 39.18%

Solution 7.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 98

Solution 8.

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Solution 9.

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Solution 10.

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Solution 11.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 20

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 20

Solution 13.

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Solution 14.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 21

Solution 15.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 37

Solution 16.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 38

Solution 17.

(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg
(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N
(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3
So, the formula is Mg3 N2

Solution 18.

Barium chloride = BaCl2.x H2O
Ba + 2Cl + x[H2 + O]
= 137+ 235.5 + x [2+16]
= [208 + 18x] contains water = 14.8% water in BaCl2.x H2O
= [208 + 18 x] 14.8/100 = 18x
= [104 + 9x] 2148=18000x
= [104+9x] 37=250x
= 3848 + 333x =2250x
1917x =3848
x = 2molecules of water

Solution 19.

Molar mass of urea; CON2H4 = 60 g
So, % of Nitrogen = 28 × 100/60 = 46.66%

Solution 20.

Element % At. mass Atomic ratio Simple ratio
C 42.1 12 3.5 1
H 6.48 1 6.48 2
O 51.42 16 3.2 1
The empirical formula is CH2O
Since the compound has 12 atoms of carbon, so the formula is
C12 H24 O12.

Solution 21.

(a) Now since the empirical formula is equal to vapour density and we know that vapour density is half of the molecular mass i.e. we have n=2 so, molecular formula is A2B4.

(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D
Empirical formula weight = V.D/3
So, n = molecular mass/ Empirical formula weight = 6
Hence, the molecular formula is A6B6

Solution 22.

Atomic ratio of N = 87.5/14 = 6.25
Atomic ratio of H= 12.5/1 = 12.5
This gives us the simplest ratio as 1:2
So, the molecular formula is NH2

Solution 23.

Element % at. mass atomic ratio simple ratio
Zn 22.65 65 0.348 1
H 4.88 1 4.88 14
S 11.15 32 0.348 1
O 61.32 16 3.83 11
Empirical formula of the given compound =ZnSH14O11
Empiricala formula mass = 65.37+32+141+11+16=287.37
Molecular mass = 287
n = Molecular mass/Empirical formula mass = 287/287=1
Molecular formula = ZnSO11H14
= ZnSO4.7H2O

Exercise 5(D)

Solution 1.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 39

Solution 2.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 40

Solution 3.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 41

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 42

Solution 5.

Molecular mass of KNO3 = 101 g
63 g of HNO3 is formed by = 101 g of KNO3
So, 126000 g of HNO3 is formed by = 126000 x 101/63 = 202 kg
Similarly,126 g of HNO3 is formed by 170 kg of NaNO3
So, smaller mass of NaNO3 is required.

Solution 6.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 43

Solution 7.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 44

Solution 8.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 45

Solution 9.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 46

Solution 10.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 47

Solution 11.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 48

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 49

Miscellaneous Exercise

Solution 1.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 50

Solution 2.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 51

Solution 3.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 52

Solution 4.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 53

Solution 5.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 54

Solution 6.

Molecular mass of urea=12 + 16+2(14+2) =60g
60g of urea contains nitrogen =28g
So, in 50g of urea, nitrogen present =23.33 g
50 kg of urea contains nitrogen=23.33kg

Solution 7.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 55

Solution 8.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 56

Solution 9.

Mass of X in the given compound =24g
Mass of oxygen in the given compound =64g
So total mass of the compound =24+64=88g
% of X in the compound = 24/88 100 = 27.3%
% of oxygen in the compound=64/88 100 =72.7%
Element % At. Mass Atomic ratio Simplest ratio
X 27.3 12 27.3/12=2.27 1
O 72.7 16 72.2/16=4.54 2
So simplest formula = XO2

Solution 10.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 57

Solution 11.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 58

Solution 12.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 59

Solution 13.

(a) Number of molecules in 100cm3 of oxygen=Y
According to Avogadros law, Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.Therefore ,number of molecules in 100 cm3 of nitrogen under the same conditions of temperature and pressure = Y
So, number of molecules in 50 cm3 of nitrogen under the same conditions of temperature and pressure =Y/100 50=Y/2

(b) (i) Empirical formula is the formula which tells about the simplest ratio of combining capacity of elements present in a compound.
(ii) The empirical formula is CH3
(iii) The empirical formula mass for CH2O = 30
V.D = 30
Molecular formula mass = V.D 2 = 60
Hence, n =mol. Formula mass/empirical formula mass= 2
So, molecular formula = (CH2O)2 = C2H4O2

Solution 14.

The relative atomic mass of Cl = (35 x 3 + 1 x 37)/4=35.5 amu

Solution 15.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 60

Solution 16.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 61

Solution 17.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 62

Solution 18.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 63
So, mass of CO2 = 22 kg
(b) According to Avogadros law ,equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
So, number of molecules of carbon dioxide in the cylinder =number of molecules of hydrogen in the cylinder=X

Solution 19.

(a) The volume occupied by 1 mole of chlorine = 22.4 litre
(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.
(c) V1/V2 = T1/T2
22.4/V2 =273/546
V2 = 44.8 litres
(d) Mass of 1 mole Cl2 gas =35.5 x 2 =71 g

Solution 20.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 64

Solution 21.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 65

Solution 22.

(a) The molecular mass of (Mg(NO3)2.6H2O = 256.4 g
% of Oxygen = 12 x 16/256
= 75%

(b) The molecular mass of boron in Na2B4O7.10H2O = 382 g
% of B = 4 x 11/382 = 11.5%

Solution 23.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 66

Solution 24.

(a) 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g
Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be = 100 x 63/252 = 25 g

(b) If 252 g of ammonium dichromate produces Cr2O3 = 152 g
So, 63 g ammonium dichromate will produce = 63 x 152/252
= 38 g

Solution 25.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 67

Solution 26.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 68

Solution 27.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 69

Solution 28.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 70
Since the pressure (760mm) remains constant , but the temperature (273+273)=546 is double, the volume of the steam also gets doubled
So,Volume of steam produced at 760mm Hg and 2730C = 4.48 × 2 = 8.96litre

Solution 29.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 71

Solution 30.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 72

Solution 31.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 73

Solution 32.

V1/V2 = n1/n2
So, no. of moles of Cl = x/2 (since V is directly proportional to n)
No. of moles of NH3 = x
No. of moles of SO2 = x/4

This is because of Avogadros law which states Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal number of molecules.

So, 20 litre nitrogen contains x molecules
So, 10 litre of chlorine will contain = x × 10/20=x/2 mols.
And 20 litre of ammonia will also contain =x molecules
And 5 litre of sulphur dioxide will contain = x × 5/20 = x/4 mols.

Solution 33.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 74

Solution 34.

(a) Volume of O2 = V
Since O2 and N2 have same no. of molecules = x
so, the volume of N2 = V
(b) 3x molecules means 3V volume of CO
(c) 32 g oxygen is contained in = 44 g of CO2
So, 8 g oxygen is contained in = 44 x 8/32 = 11 g
(d) Avogadro’s law is used in the above questions.

Solution 35.

(a) 444 g is the molecular formula of (NH4)2 PtCl6
% of Pt = (195/444) x 100 = 43.91% or 44%

(b) simple ratio of Na = 42.1/23 = 1.83 = 3
simple ratio of P = 18.9/31 = 0.609 = 1
simple ratio of O = 39/16 = 2.43 = 4
So, the empirical formula is Na3PO4

Solution 36.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 75

Solution 37.

According to Avogadros law:
Equal volumes of all gases, under similar conditions of temperature and pressure ,contain equal number of molecules.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 76

Solution 38.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 77

Solution 39.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 78

Solution 40.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 79

Solution 41.

(i) D contains the maximum number of molecules because volume is directly proportional to the number of molecules.

(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.
V1/V2 = n1/n2
V1/V2 = n1/2n1
So, V2 = 2V1

(iii) Gay lussac’s law of combining volume is being observed.

(iv) The volume of D = 5.6 4 = 22.4 dm3, so the number of molecules = 6 x 1023 because according to mole concept 22.4 litre volume at STP has = 6 x 10 23 molecules

(v) No. of moles of D = 1 because volume is 22.4 litre
so, mass of N2O = 1 44 = 44 g

Solution 42.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 80

Solution 43.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 81

Solution 44.

(a) Element % Atomic mass Atomic ratio Simple ratio
K 47.9 39 1.22 2
Be 5.5 9 0.6 1
F 46.6 19 2.45 4
so, empirical formula is K2BeF4

(b) 3CuO + 2NH3 → 3Cu + 3H2O + N2
3 V 2 V 3 V 1V
3 x 80 g of CuO reacts with = 2 x 22.4 litre of NH3
so, 120 g of CuO will react with = 2x 22.4 x 120/80 x 3
= 22.4 litres

Solution 45.

(a) The molecular mass of ethylene(C2H4) is 28 g
No. of moles = 1.4/28 = 0.05 moles
No. of molecules = 6.023 x1023 x 0.05 = 3 x 1022 molecules
Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 X V.D
S0, V.D = 28/2 = 14

Solution 46.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 82

Solution 47.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 83
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 84

Solution 48.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 85
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 86
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 87

Solution 49.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 88
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 89

Solution 50.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 90
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 91

Solution 51.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 92
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 93

Solution 52.
Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 94

Solution 42.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 95

Solution 48.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 97

Solution 49.

Selina Concise Chemistry Class 10 ICSE Solutions Mole Concept and Stoichiometry img 97

Solution 50.

(a) (i) element % atomic mass at. ratio simple ratio
C 14.4 12 1.2 1
H 1.2 1 1.2 1
Cl 84.5 35.5 2.38 2
Empirical formula = CHCl2
(ii) Empirical formula mass = 12+1+71= 84 g
Since molecular mass = 168 so, n = 2
so, molecular formula = (CHCl2)2 = C2H2Cl4

(b) (i) C + 2H2SO4 → CO2 + 2H2O + 2SO2
1 V 2 V 1 V 2 V
196 g of H2SO4 is required to oxidized = 12 g C
So, 49 g will be required to oxidise = 49 x 12/196 = 3 g
(ii) 196 g of H2SO4 occupies volume = 2 x 22.4 litres
So, 49 g H2SO4 will occupy = 2 x 22.4 x 49/196 = 11.2 litre
i.e. volume of SO2 = 11.2 litre

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Selina Concise Chemistry Class 10 ICSE Solutions Electrolysis

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Selina Concise Chemistry Class 10 ICSE Solutions Electrolysis

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 6 Electrolysis. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 6 Electrolysis

Exercise Intext 1

Solution 1.

(a) Powdered sodium chloride (common salt) does not conduct an electric current, but it does so when dissolved in water or when melted.

(b) Molten lead bromide conducts electricity .It is called an electrolyte. It is composed of lead ions and bromide ions. The lead ions are positivelycharged and are called cations. The bromide ions are negatively charged and are called anions.

(c) Substances which conduct electricity in the solid state are generally metals.

(d) The electron releasing tendency of zinc is more than that of copper.

(e) A solution of HCl gas in water conducts electricity because it ionizes, but a solution of HCl gas in toluene does not conduct an electric current because it does not ionize in toluene.

Solution 2.

(a) Electrolysis: It is the process of decomposition of a chemical compound in aqueous solutions or in molten state accompanied by a chemical change using direct electric current.

(b) Non-electrolyte: It is a compound which neither in solution nor in the molten state allows an electric current to pass through it.

(c) Cation and anion: Atoms which carry positive charge are called cations.
Atoms which carry negative charge are called anions.

(d) Weak electrolyte: Electrolytes which allow small amount of electricity to flow through them and are partially dissociated in fused or aqueous solution are called weak electrolyte.

Solution 3.

(a) Difference between Modern explanation and Arrhenius explanation for the theory of electrolysis:
Arrhenius considered that water ionizes electrolytes but Modern theory explained that electrolytes are ionic even in solid state and their ions are held by strong electrostatic forces which make them immobile. Water renders these ions mobility by breaking the electrostatic forces.

(b) Difference between electrolytic dissociation and ionization :

IonisationDissociation
1. Formation of positively or negatively charged ions from molecules which are not initially in the ionic state.1. Separation of ions which are already present in an ionic compound.
2. Polar covalent compounds show ionization. e.g. HCl, H2CO3, NH4OH etc.1. Electrovalent compounds show dissociation. e.g. Potassium chloride , lead bromide, etc.

(c) A cation and anion:

CationAnion
1. Are positively charged ions.Are negatively charged ions.
2. Migrate to cathode during electrolysis.Migrate to anode during electrolysis.
3. Gain electron from the cathode and get reduced to become a neutral atom.Lose electrons to the anode and get oxidized to become a neutral atom.

(d) Electrolytic dissociation and thermal dissociation:
Electrolytic dissociation is the dissociation of an electrovalent compound into ions in the fused state or in aqueous solution state.

Thermal dissociation: Reversible breakdown of a chemical compound into simpler substances by heating it. The splitting of ammonium chloride into ammonia and hydrogen chloride is an example. On cooling, they recombine to form the salt.

Solution 4.

(a) Sodium carbonate
(b) NH4OH
(c) An inert electrode: graphite and Active electrode: silver
(d) H+
(e) Electrode is cathode
(f) Graphite

Solution 5.

Electrolysis is a redox process. The reaction at the cathode involves reduction of cations as they gain of electrons while the reaction at anode involves oxidation of anions as they loss of electrons to become neutral.
Example: Dissociation of sodium chloride during electrolysis.

Cathode : Na+ + e– → Na (reduction)
Cl– – e– → Cl (oxidation)
Cl + Cl → Cl2
Overall reaction: 2NaCl → 2Na + Cl2

Exercise Intext 2

Solution 1.

(a) Glucose, Kerosene
(b) NaCl and NaOH
(c) CH3COOH and NH4OH

Solution 2.

(a) Cane sugar is a compound which does not have ions even in solution and contains only molecules. Hence, it does not conduct electricity. On the other hand, sodium chloride solution contains free mobile ions and allows electric current to pass through it. This makes it a good conductor of electricity.

(b) Hydrochloric acid is a strong electrolyte and dissociates completely in aqueous solution. The solution contains free mobile ions which allow electric current to pass through it. Hence, hydrochloric acid is a good conductor of electricity.

(c) Hydrogen is placed lower in the electrochemical series and sodium is placed at a higher position. This is because H+ ions are discharged more easily at the cathode than Na+ during electrolysis and gains electrons more easily.
Therefore, H+ ion is reduced at the cathode and not Na+ ion.

Solution 3.

(a) Zn occurs readily as ion whereas Cu occurs more readily as metal in nature.

(b) Copper is above silver in the electrochemical series and is thus more reactive than silver. So, copper displaces silver from silver nitrate. Hence, we cannot store AgNO3 solution in copper vessel.
Cu +AgNO3 → Cu(NO3)2 + 2Ag

(c) Copper is more active than Ag.

Solution 4.

(a) By treating its salt with a more reactive metal.
(b) By supplying two electrons to Cu+2
Cu+2 + 2e– → Cu

Solution 5.

In the aqueous state, the slightly negatively charged oxygen atoms of the polar water molecule exerts a pull on the positively charged sodium ions. A similar pull is exerted by the slightly charged hydrogen atoms of the water on the negatively charged chloride ions. Thus the ions become free in solution. These free ions conduct electricity.
In the molten state, the high temperatures required to melt the solid weakens the bond between the particles and the ions are set free.

Solution 6.

(a) Two anions are and OH–.

(b) OH– is discharged at anode and the main product of the discharge of OH– is O2
Reaction is :
OH– → OH + e–
4OH → 2H2O + O2

(c) The product formed at cathode is hydrogen. The reaction is :
H+ + e– → H
H + H → H2

(d) No change in colour is observed.

(e) Dilute sulphuric acid catalyse the dissociation of water molecules into ions, hence electrolysis of acidified water is considered as an example of catalysis.

Solution 7.

(a) Labelled diagram of electrolytic cell is:

(b) The ions present in the cell are Cu2+, H+, SO42- , OH–.
(c) SO42- and OH– ions both migrate towards anode.
(d) Both Cu2+ and H+ ions migrate towards cathode.
(e) SO42- and H+ will not discharge at electrodes.
(f) Reaction at cathode:
Cu+2 +2e– → Cu
(g) Reaction at anode:
OH– – e– → OH
2OH + 2OH → 2H2O + O2
(h) Sulphate ions are the spectator ions because they do not change in the reaction.

Solution 8.

(a) Reaction at anode during the electrolysis ofvery dilute sulphuric acid:
OH– → OH + e–
4OH → 2H2O + O2

(b) Reaction at anode during the electrolysis of aqueous copper sulphate solution
4OH– → 4OH + 4e–
4OH → 2H2O + O2

(c) Reaction at anode during the electrolysis of sodium chloride solution
2Cl– → Cl2 + 2e–

(d) Reaction at anode during the electrolysis of fused lead bromide
Br– – e– → Br
Br + Br → Br2

(e) Reaction at anode during the electrolysis of magnesium chloride (molten)
2Cl– → Cl2 +2e–

(f) Concentrated HCl,
HCl in the pure liquid state is unionised and hence does not conduct electricity.

(g) Very dilute HCl,
Cl– – e– → Cl
Cl  + Cl  → Cl2

Solution 9.

(a) Electrolyte
(b) Nickel
(c) Cathode
(d) Anode
(e) Cations

Exercise 1

Solution 1.

(a) During electrolysis of lead bromide, there is loss of electrons at anode by bromine and gain of electrons at cathode by lead. Thus oxidation and reduction go side by side. Therefore, it is a redox reaction.
PbBr2 → Pb+2 + 2Br–

(b) The blue colour of copper ions fades due to decrease in Cu+2 ions and finally the solution becomes colourless as soon as Cu+2 ions are finished.

(c) Lead bromide dissociate into ions in the molten state whereas it does not dissociate in solid state. The ions become free when lead bromide is in molten state but in the solid state the ions are not free since they are packed tightly together due to electrostatic force between them. Therefore, lead bromide undergoes electrolytic dissociation in the molten state.

(d) Aluminium has great affinity towards oxygen, so it is not reduced by reducing agent. Therefore it is extracted from its oxide by electrolytic reduction.

(e) As per electrolytic reactions, 4H+1 are needed at cathode and 4OH– at the anode and two molecules of water are produced at the anode. Hence for every two molecules of water, two molecules of hydrogen and one molecule of oxygen are liberated at the cathode and anode respectively.

(f) This is because HNO3 is volatile.

(g) Ammonia is a covalent compound. Therefore, it is unionized in the gaseous state but in the aqueous solution it gives NH4OH which is a weak electrolyte and dissociates into ions.

(h) Graphite is unaffected by the bromine vapours.

(i) Silver nitrate is not used as electrolyte for electroplating with silver because the deposition of silver will be very fast and hence not very smooth and uniform.

(j) Carbon tetrachloride is a liquid and does not conduct electricity because it is a covalent compound and there are no free ions present and contain only molecules.

Solution 2.

(a) Strong electrolyte : Dilute hydrochloric acid, dilute sulphuric acid, ammonium chloride, sodium acetate
(b) Weak electrolyte: Acetic acid, ammonium hydroxide
(c) Non-electrolyte: Carbon tetrachloride

Solution 3.

(a) Molecules
(b) Will not

Solution 4.

Water is a non-conductor of electricity and consists entirely of molecules. It can be electrolytically decomposed by addition of traces of dilutesulphuric acid which dissociate as H+ and SO42- ions and help in dissociating water into H+ and OH–, water being a polar solvent.

Solution 5.

AnodeElectrolyteCathode
Silver plating of spoonPlate of pure clean silverSolution of potassiumargentocyanideArticle to be electroplated
Purification of copperImpure copperSolution of coppersulphate and dilutesulphuric acidThin strip of pure copper

Solution 6.

Electricity, Chemical

Solution 7.

(b) CuSO4 is preferred as an electrolyte.

(c) The copper anode continuously dissolves as ions in solution and is replaced periodically. The electrolyte dissociates into Cu+2 ions which migrate towards the iron object taken as the cathode and are deposited as neutral copper atoms on the cathode.
Electrolyte: Aqueous solution of nickel sulphate
Dissociation: CuSO4 → Cu2+ + SO42-
H2O → H+ + OH–
Electrodes:
Cathode: Article to be electroplated
Anode: Block of pure copper
Electrode reactions:
Reaction at cathode: Cu2+ + 2e–→ Cu (deposited)
Reaction at anode: Cu – 2e–→ Cu2+

Solution 1 (2004).

(a) X → X2+ + 2e– , Y + 3e– → Y3-
(b) Y2 + 3X → X3Y2
(c) (i) It is used for the electroplating of metals.
(ii) It is also used in purification of metals.
(d) Cathode, Anode

Solution 1 (2005).

(a) Because Copper is an electronic conductor as it is a metal.
(b) In solid sodium chloride, Na+ and Cl – ions are not free due to strong electrostatic forces of attraction among them. The ions, therefore are unable to move to any large extent when electric field is affected. Hence no current.

Solution 1 (2006).

(a) (i) The name of electrode A is Platinum anode and that of electrode B is platinum or copper cathode.
(ii) Anode act as oxidizing electrode.
(b) AgNO3 solution will turn blue.

Solution 1 (2007).

(i) Molten ionic compound: Strong electrolytes
(ii) Carbon tetrachloride: Non-electrolyte
(iii) An aluminium wire: Metallic conductor
(iv) A solution containing solvent molecules, solute molecules and ions formed by dissociation of solute molecules: Weak electrolyte
(v) A sugar solution with sugar molecules and water molecules: Non- electrolyte

Solution 1 (2008).

(a) The reaction takes place at anode. This is an example of oxidation.
(b) Cu+2 will discharge easily at cathode.
Reaction at cathode:
Cu+2 +2e– → Cu
(c) Carbon tetrachloride is a non-electrolyte because it is a covalent compound. It does not ionize and hence do not conduct electricity.

Solution 2 (2004).

(a) Non-electrolyte contains molecules.
(b) Molecules of HX and H+ and X– ions.
(c) Loss
(d) The electrolyte used for the purpose must contain the ions of metal which is to be electroplated on the article.
(e) The reaction at the cathode involves reduction of cations as they gain electrons to become neutral atoms while that at anode involves oxidation of anions as they lose electrons to become neutral.
Example: Dissociation of sodium chloride during electrolysis.
NaCl → Na+ + Cl–
At cathode: Na+ + e– Na (Reduction)
At anode: Cl– – e– → Cl(oxidation)
Cl + Cl → Cl2
Overall reaction: 2NaCl → 2Na + Cl2

Solution 2 (2005).

Hydrogen gas is released at cathode when acidulated water is electrolyzed.

Solution 2 (2008).

During the electrolysis of molten lead bromide. Lead is deposited at cathode.

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Selina Concise Chemistry Class 10 ICSE Solutions Metallurgy

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Selina Concise Chemistry Class 10 ICSE Solutions Metallurgy

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 7 Metallurgy. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Chemistry Chapter 7 Metallurgy

Exercise 7(A)

Solution 1.

Three classes in which elements are classified are:
Metals, Non-metals and Metalloids
Copper was the first metal used by man.

Solution 2.

(a) The metal which is a constituent of blood pigment is Iron (Fe)
(b) The metal which is a constituent of plant pigment is Magnesium (Mg).

Solution 3.

(a) Nitrogen: It is used to preserve food.
(b) Hydrogen: It is used in the hydrogenation of vegetable oils to make ghee.
(c) Carbon: It is essential for the growth and development of living beings.

Solution 4.

The metal which is present in abundance in earths crust is aluminium.
The non-metal which is present in abundance in the earth crust is oxygen.

Solution 5.

Metals are defined as the elements which form positive ions by the loss of electrons.
Non-metals are the elements which form negative ions by the gain of electrons.

Solution 6.

(a) Alkali metals: They are placed in IA group, the first column on the left of the periodic table.
(b) Alkaline earth metal: They are placed in IIA group, the second column on the left of the periodic table.
(c) Iron and Zinc: Fe is placed in VIII group and Cu is placed in IB group.
(d) Aluminium: It is placed in IIIA group present on the right of periodic table.

Solution 7.

(a) Alkali metals:-
(i) Bonding: All alkali metal salts are ionic in nature.
(ii) Action of air: The react rapidly with oxygen and water vapour in the air.
(iii) Action of water: They react violently with water and produce hydrogen gas.
2M + 2H2O → 2MOH + H2
(iv) Action of acid: They react violently with dil. HCl and dil. H2SO4 to produce hydrogen gas.
2M + 2HCl → 2MCl + H2

(b) Alkaline earth metal:-
(i) Bonding: All alkaline earth metal salts except beryllium are ionic compounds.
(ii) Action of air: They are less reactive than alkali metals.
(iii) Action of water: They react with water to produce hydrogen gas.
M + 2H2O → M(OH)2 + H2
(iv) Action of acid: They react with dilute HCl and dil. H2SO4 to produce hydrogen gas.
M + 2HCl → MCl2 + H2

Solution 8.

Elements which show properties of both metals and non-metals are called metalloids.
For example: Silicon, Germanium.

Solution 9.

Hydrogen is placed with alkali metals as it has one electron similar to the alkali metals.

Solution 10.

(a) Bromine
(b) Lead
(c) Gallium
(d) Carbon
(e) Sodium
(f) Sodium
(g) Tungsten
(h) Carbon fibre
(i) Carbon
(j) Mercury

Solution 11.

(i) Ion formation: Metals form positive ions by loss of electrons whereas non-metals form negative ions by gain of electrons.

(ii) Discharge of ions: Metals are discharged at the cathode during electrolysis whereas non-metals are liberated at the anode during electrolysis.

(iii) Nature of oxide formed: Oxides of metals are usually basic. Soluble basic oxides dissolve in water forming an alkaline solution whereas oxides of non-metals are usually acidic. Soluble acidic oxides dissolve in water forming an acidic solution.

(iv) Oxidizing and reducing property: Metals ionize by loss of electrons and hence are reducing agents whereas non-metals ionize by gain of electrons and hence are oxidizing agents.

(v) Reaction with acids: Metals above hydrogen in activity series usually replace hydrogen from dilute non-oxidising acids whereas non-metals do not react with dilute hydrochloric acid or sulphuric acid.

Solution 12.

(a) Na – e–  → Na+
(b) N + 3e– → N3-
(c) Cl + e– → Cl–
(d) Mg – 2e– → Mg2+
(e) M + 2HCl → MCl2 + H2
(f) Mg + H2SO4 → MgSO4 + H2

Solution 13.

(a) Fe2O3
(b) PbO
(c) Mn2O7
(d) NO

Solution 14.

Exercise 7(B)

Solution 1.

(a) Mercury and gallium
(b) Sodium and potassium
(c) Mercury
(d) Iodine
(e) Graphite
(f) Zinc
(g) Neon , Argon
(h) CrO3 , Mn2O7
(i) Al2O3,PbO
(j) Potassium , sodium
(k) Basic copper(II) sulphate
(l) Aluminium , Oxygen
(m) Hydrogen
(n) Carbon
(o) Iron

Solution 2.

(a) Occurrence of metals: The metals placed at the top of activity series are most reactive, so they always exist in the combined state whereas the metals placed below the activity series are least reactive, so they can be found in the isolated state also.

(b) Tendency to corrosion: The metals lying above the hydrogen in activity series can easily react with moisture and air and corrode easily whereas the metals such as gold and platinum do not corrode easily.

(c) Reaction with water: The ability of the metals to reduce water to hydrogen decreases on moving down the series.

Potassium and sodium reacts with cold water whereas magnesium reacts with warm water and aluminium, zinc and iron reacts with steam.

(d) Reaction with acids: All the metals above hydrogen, in the activity series, reduce hydrogen ions from dil. hydrochloric or sulphuric acid and give out hydrogen gas. The rate of reaction decreases on moving down the series.

Solution 3.

selina-icse-solutions-class-10-chemistry-metallurgy-7b-3

Solution 4.

The metals placed higher in the activity series (i.e. Na and K) are stable to heat and soluble in water.
Whereas metals like Ca, Mg, Al, Zn, Fe, Pb, Cu decompose on heating with decreasing vigour to form metal oxide and carbon dioxide.
The metals which lie below in the activity series (i.e. Hg, Ag) decompose on heating to form metal, oxygen and carbon dioxide.

Solution 5.

(a) Alkali metals like sodium and potassium are kept in kerosene as they react with moisture and air.
(b) (i) Basic lead carbonate is a mixture of lead hydroxide and lead carbonate.
(ii) Brown powder is mainly hydrated iron(III) oxide (Fe2O3.xH2O)

Solution 6.

Oxides of metals like Na, K, Ca, Mg, Al are stable to heat and so can be reduced only by electrolysis.
Zinc oxide can be reduced by coke only.
Oxides of iron, lead and copper are reduced by C, CO, H2 and NH3.
Oxides of mercury and silver decompose to give metal and oxygen.

Solution 7.

Metal A is more reactive than Metal B.
(a) Metal A is Na (Sodium). Metal B is Ca (Calcium).
Reaction with HCl:

selina-icse-solutions-class-10-chemistry-metallurgy-7b-7

(b) (i) Oxides: Sodium and calcium oxides are stable to heat.
(ii) Hydroxides: Sodium hydroxide is stable to heat whereas calcium hydroxide decomposes on heating to metal oxide and water vapour.
(iii) Carbonates: Sodium carbonate is stable to heat whereas calcium carbonates decompose on heating to form calcium oxide and carbon dioxide.
(iv) Nitrates: Sodium nitrate on heating form nitrite and oxygen whereas calcium nitrate decomposes on heating to form calcium oxide, nitrogen dioxide and oxygen.

Solution 8.

(a)

MetalsNon-metals

(i) Good conductors of heat
(ii) Malleable
(iii) Form positive ions
(iv) Form basic oxides

Poor conductors of heat
Non-Malleable
Forms negative ions
Form acidic oxides

(b) Valence electrons present in :
(i) Metals have 1, 2 or 3 valence electrons.
(ii) Non-metals have 5, 6 or 7 valence electrons.

Solution 9.

When the surface of metal is attacked by air, moisture or any other substance around it, the metal is said to corrode and the phenomenon is known as corrosion.
Necessary conditions for corrosion are:

  1. Presence of oxygen and moisture.
  2. Metals which are placed higher in the activity series corrode more easily.

Solution 10.

Conditions for increase of corrosion are:

  1. Presence of oxygen and moisture.
  2. Metals which are placed higher in activity series corrode more easily
  3. Dissolved salts in water act as electrolyte and enhance the rate of corrosion.
  4. The presence of pollutants like NO2and CO2increases rusting.

Solution 11.

Corrosion of metals is an advantage as it prevents the metal underneath from further damage. For example: On exposure to air, the surface of metal like aluminium and Zinc forms layers of their oxides which are very sticky and impervious in nature and hence act as protective layer. This layer protects the metal from further damage.

Solution 12.

Rusting is the slow oxidation of iron by atmospheric oxygen in the presence of water.
Equation:
4Fe + 3O2 + 2x H2O → 2Fe2O3.xH2O

Solution 13.

Two conditions necessary for rusting of iron are:

  1. Air
  2. Water

Solution 14.

By painting an iron object, the iron do not come in contact with atmospheric reagents .This prevents rusting.

Solution 15.

Galvanisation is the process of applying a protective zinc coating to steel or iron, in order to prevent rusting.
The zinc coating does not allow iron to come in contact with air and moisture and thus protects it from rusting.

Solution 16.

Silver gets tarnished when exposed to the atmosphere which contains pollutant H2S and forms a black coating of Ag2S.
Copper forms a green deposit on its surface when exposed to moist air. This is usually basic copper (II) sulphate.

Solution 17.

Aluminium forms white colour oxide on exposure to the atmosphere. This white colour oxide prevents it from further corrosion whereas iron reacts with air to form hydrated oxide called rust. So, iron undergoes corrosion to greater extent.

Solution 18.

The noble metals such as gold and platinum do not corrode easily.

Solution 19.

Gold is the most unreactive metal so it does not react with air or water and other gases in atmosphere. So gold does not corrode. That is why gold look new after several years of use.

Exercise 7(C)

Solution 1.

The process used for the extraction of metals in their pure form from their ores is referred to as Metallurgy.
The processes involved in Metallurgy are:

  1. Crushing and Grinding
  2. Concentration
  3. Roasting andcalcination
  4. Reduction
  5. Refining

Solution 2.

(a) A metal which occurs as sulphide is lead.
(b) A metal which occurs as halide is silver.
(c) A metal which occurs as carbonate is zinc.
(d) A metal which occurs as oxide is iron.

Solution 3.

(a) Minerals are naturally occurring compounds of metals which are generally present with other matter such as soil, sand, limestone and rocks. Ores are those minerals from which the metals are extracted commercially at low cost and comfortably. All ores are minerals, but all minerals are not necessarily ores.

(b) Ores are those minerals from which the metals are extracted commercially at low cost and with minimum effort. A metallic compound is a compound that contains one or more metal elements. Examples: AgNO3 – Silver nitrate is a metallic compound.

Solution 4.

The metals that can be extracted from the following ores are:
(a) Bauxite- Aluminium
(b) Calamine- Zinc
(c) Haematite- Iron

Solution 5.

Three objectives achieved during the roasting of ores is:

  1. It removes moisture from ores.
  2. It makes the ore porous and more reactive.
  3. It expels volatile impurities.
  4. It convertssulphideores into oxides.

Solution 6.

(a) Hydraulic washing: The difference in the densities of the ore and the gangue is the main criterion.
(b) Forth floatation: This process depends on the preferential wettability of the ore with oil and the gangue particles by water.
(c) Electromagnetic separation: Magnetic properties of the ores.

Solution 7.

(a) The processes involved in

(i) Processes involved in concentration are:

  1. Hydrolytic method
  2. Magnetic Separation
  3. Froth floatation
  4. Leaching

(ii) Processes involved in Refining of ores are:

  1. Distillation
  2. Liquation
  3. Oxidation
  4. Electro- refining

(b) Potassium and sodium oxides cannot be reduced by carbon, carbon monoxide and hydrogen.

Solution 8.

(a) Flux: A flux is a substance that is added to the charge in a furnace to remove the gangue.
(b) Gangue: Earthly impurities including silica, mud etc., associated with the ore are called gangue.
(c) Slag: It is the fusible product formed when flux reacts with impurities during the extraction of metals.
(d) Smelting: Smelting is the process of reducing the roasted oxide ore and removing the gangue with the help of an appropriate flux added with the ore.

Solution 9.

Iron and zinc are quite reactive and hence they do not occur in the free state. The compounds of metals found in nature are their oxides, carbonate and sulphides.

Solution 10.
selina-icse-solutions-class-10-chemistry-metallurgy-7c-10

Solution 11.
selina-icse-solutions-class-10-chemistry-metallurgy-7c-11

Solution 12.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-12

Solution 13.

Oxides of highly active metals like potassium, sodium, calcium, magnesium and aluminium have great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide or hydrogen.

Metals in the middle of activity series (iron, zinc, lead, copper) are moderately reactive and are not found in oxide form. These are found in nature as sulphides or carbonate. These are first converted into oxides and can be reduced by C, CO or H2.
selina-icse-solutions-class-10-chemistry-metallurgy-7c-13
Metals low in the activity series is very less reactive and oxides of these metals are reduced to metals by heating alone.

Solution 14.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-14

Solution 15.

Aluminium has a great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide. So it is extracted from its oxide by electrolysis.
Metals like copper, lead and iron are placed in the middle of the activity series and re moderately reactive and their oxides can be reduced by carbon, CO and hydrogen.
Mercury and silver are less reactive and are placed lower in the reactivity series. The oxides of these metals are reduced to metals by heating their oxides.

Solution 16.

The process used for the concentration of the ore is froth floatation process.

Solution 17.

(a) The purification depends upon:

  1. Nature of metal.
  2. Nature of impurities present in the metal.
  3. Purpose for which metal is to be used

(b) Methods used for purification are:

  1. Distillation
  2. Liquation
  3. Oxidation
  4. Electro-refining

(c) The impure metal is made the anode, while a thin sheet of pure metal is made the cathode. Electrolyte used is a salt solution of a metal which is to be refined. Pure metal deposits at the cathode and impurities settle down forming anode mud.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-17

Solution 18.

selina-icse-solutions-class-10-chemistry-metallurgy-7c-18

Exercise 7(D)

Solution 1.

Position in the Periodic Table: Period 3,Group IIIA(13)

Solution 2.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-2

Solution 3.

Bauxite ore contains approximately 60% aluminium oxide. The rest being sand, ferric oxide and titanium oxide.

Solution 4.

Red mud consists of ferric oxide, sand etc. left after bauxite dissolves in NaOH forming sodium aluminate and is removed by filtration.

Solution 5.

As aluminium has great affinity for oxygen, so it is stable compound. It is not easily reduced by common reducing agents like carbon, carbon monoxide or hydrogen. Hence, electrolytic reduction is chosen as the method for reducing alumina.

Solution 6.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-6

Solution 7.

(a) The process by which refining of aluminium is done is called Hoope’s electrolytic process.
(b) Molten impure aluminium forms the bottom layer. The bottom layer has carbon lining and serves as anode.
Pure molten aluminium with carbon electrodes serves as cathode in top layer.
(c) Reactions at the two electrodes are:
Anode: Al -3e– → Al3+
Cathode: Al3+ + 3e– → Al

Solution 8.

Reaction of aluminium:

(a) Air: Aluminium forms oxide at room temperature.
Aluminium powder burns in air at about 8000C forming its oxide and nitride with a bright light.
4Al + 3O2 → 2Al2O3
2Al + N2 → 2AlN

(b) Water: Water has no action on aluminium due to layer of oxide on it.
When steam is passed over pure heated aluminium, hydrogen is produced.
2Al + 3H2O → Al2O3 + 3H2

(c) Acid: It reacts with acids to produce salt and hydrogen.
2Al + 6HCl → 2AlCl3 + 3H2
Dilute sulphuric acid reacts with metal to liberate hydrogen.
2Al + 3H2SO4 (dilute) → Al2(SO4)3 + 3H2

Concentrated sulphuric acid reacts with aluminium to produce sulphur dioxide.
2Al + 6H2SO4 → Al2(SO4)3 + 6H2O + 3SO2
Dilute and concentrated nitric acid does not attack the metal aluminium.

(d) Base: Aluminium reacts with boiling and dilute alkalies to produce meta aluminate while with fused alkali produce aluminate.
2Al+ 2NaOH +2H2O → 2NaAlO2 +3H2
(Sodium meta aluminate)
2Al + 6NaOH → 2NaAlO3 +3H2
(Sodium aluminate)

Solution 9.

The role of cryolite in the electrolytic reduction of alumina in Hall’s process is :

  1. Lowers the fusion temperature from 20500C to 9500C and enhances conductivity.
  2. Increases its conductivity since pure alumina is almost a non-conductor of electricity.
  3. Cryoliteacts as a solvent for the electrolytic mixture.

Solution 10.

(a) Aluminium is more active metal but it gets oxidized and forms a thin protective layer on its surface which prevents further corrosion.
(b) Aluminium vessels should not be cleaned with powders containing alkalis because it results in the formation of meta aluminates and hydrogen.
2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

Solution 11.

selina-icse-solutions-class-10-chemistry-metallurgy-7d-11-1
selina-icse-solutions-class-10-chemistry-metallurgy-7d-11-2
selina-icse-solutions-class-10-chemistry-metallurgy-7d-11-3
(c) A layer of aluminium is formed on iron at high temperature during cooking and food becomes deficient in iron.

Solution 12.

(a) In the electrolytic reduction of alumina, the graphite (anode) is oxidized by oxygen to CO and further forms CO2, so it is consumed and has to be replaced from time to time.
2C + O2 → 2CO
2CO + O2 → 2CO2

(b) Roasting provides oxygen to convert metallic sulphides into metallic oxide and SO2 which takes place when heated in excess of air.
Carbonate is converted into oxide by loss of CO2 which takes place in the absence of air and when heated strongly.

(c) Aluminium has a great affinity towards oxygen and so cannot be reduced by carbon or carbon monoxide or hydrogen whereas lead oxide can be easily reduced to metal lead by carbon.
PbO + C → Pb + CO

Solution 13.

(a) Flux combines with the gangue to form a fusible mass called slag.
(b) It forms slag[CaSiO3] with silica.
(c) It is removed from upper outlet, slag being lighter float on molten iron.

Solution 14.

(a) Froth flotation process: Zinc blende[ZnS]
(b) Magnetic Separation: Haematite[Fe2O3]

Solution 15.

Electrolytic Reduction

  1. It is removal of oxide or halide from a metal.
  2. Oxides of highly active metals like Na,K,Ca,Mg,Al are reduced by electrolytic reduction of their fused salts.
  3. Oxides of these metals have great affinity for oxygen than carbon and cannot be reduced by carbon or CO or hydrogen.

Electrolytic refining of metals is the separation of residual impurities like Si and phosphorus.

  1. Presence of other metals and non-metals like Si and phosphorus.
  2. Unreduced oxides and sulphides of metals.

It depends upon:

  1. Nature of metal
  2. Purpose for which metal is to be obtained.
  3. Nature of impurities present.
  4. Impure metal is made anode while a thin sheet of pure metal is made cathode and electrolyte used is a salt of solution of a metal to be refined.

Solution 16.

The three ways in which metal zinc differs from the non-metal carbon is:

  1. Zinc has avalency2 and carbon has valency 4.
  2. Zinc does not form hydride but carbon does (CH4).
  3. Oxides of zinc areamphoteric(ZnO) whereas oxides of carbon are acidic (CO2) and neutral (CO).

Exercise 7(E)

Solution 1.

(a) To prevent from rusting.
(b) Due to strong electropositive nature, it easily forms Zn+2 ions.
(c) Antiseptic in face creams.

Solution 2.

(a) Aluminium:

  1. Being a strong, light and corrosion resistant metal, it is used in alloys.
  2. Aluminium is light, it has high tensile strength, is resistant to corrosion, good conductor of heat, unaffected by organic acids and has attractive appearance. So it is used for making cooking utensils, in building and construction work.
  3. Aluminium has a strong affinity for oxygen so it is used as a deoxidizer in the manufacture of steel.

(b) Zinc:

  1. Zinc has a strong electropositive character, so it is used for coating iron and steel sheets to prevent them from rusting and this process is known as galvanization.
  2. Due to strong electropositive nature, it forms Zn+2ions, so it is used to make dry cell containers which act as negative electrode.
  3. Zinc act as a reducing agent for many organic reductions and these reductions are employed in manufacturing drugs, dyes.

Solution 3.

(a) Zinc is electropositive metal than iron, gets oxidized and saves iron. Also zinc forms protective layer of ZnO on iron. This layer is sticky and impervious in nature and protects the iron metal underneath from rusting.

(b) A neutral gas other than oxygen which is formed at anode during electrolysis of fused alumina is carbon monoxide.

(c) Nitric acid can be stored in aluminium containers as the dilute and conc. nitric acid does not react with aluminum. It renders aluminium passive due to the formation of an oxide film on its surface.

Solution 4.

(a) Cast iron: It is used in drain pipes, gutter covers, weights and railings.
(b) Wrought iron: It is used in chains, horse shoes and electromagnets.
(c) Mild steel: It is to manufacture nuts, bolts etc.
(d) Hard steel: It is used to make tools.

Solution 5.

(a) Galvanized iron sheets
(b) Zinc
(c) Zinc

Solution 6.

(a) Aluminium being strong, light and corrosion resistant metal is used for making alloy.
(b) Aluminium is light, malleable and does not rust so it is used for wrapping chocolates.
(c) To prevent them from rusting.
(d) It is used in aluminothermy as it is a good reducing agent.
(e) As aluminium forms a film of aluminium oxide, it protects the ships from corrosion. So it is used for making ships.

Solution 7.

(a) A mixture of 3 parts of ferric oxide (Fe2O3) and one part of aluminium powder (Al).
(b) A mixture of Potassium chlorate and magnesium powder is the ignition mixture.
(c) Fe2O3 + 2Al → Al2O3 + 2Fe + heat

Solution 8.

Alloy is a homogeneous mixture of two or more metals or of one or more metals with certain non-metallic elements.
The properties of alloys are often greatly different from those of the components.
For example: Gold is too soft to be used without small percentage of copper.
A low percentage of molybdenum improves the toughness and wear resistance of steel.
Bell metal is more sonorous than copper or tin.
Alnico an alloy of aluminium, nickel and cobalt can lift 60 times its own mass.
These added elements improve hardness, wear resistance, toughness and other properties.

Solution 9.

Alloy’s nameCompositionUses
1. Stainless steel73% Fe, 18%Cr,8%Ni,1%CUsed for making utensils, cutlery, ornamental pieces and surgical instruments.
2. Manganese steel85% Fe,1%C ,14%MnUsed for making rock drills and armour plates.
3. Tungsten steel84%Fe, 5%W, 1%CUsed for cutting tools for high speed lathes.

Solution 10.

The other element in Brass is Zinc.
The other elements in Bronze are Tin and Zinc.

Solution 11.

(a) Duralumin
(b) Solder
(c) Brass
(d) Zinc amalgam

Solution 12.

A mixture or an alloy of mercury with a number of metals or an alloy such as sodium, zinc, gold and silver as well as with some non-metals is known as amalgam.
Dental amalgam is a mixture of mercury and a silver tin alloy.

Solution 13.

(a) Two properties of brass that make it more useful than its components are:

  1. It is malleable and ductile.
  2. It resists corrosion.
  3. Can be easily cast.

(b) A metal which forms a liquid alloy at ordinary temperature is sodium.

Solution 14.

Magnalium is an alloy of aluminium with composition 90-95% and magnesium with composition 10-5%. It is used for making aircrafts.

Solution 15.

The constituents of
(a) Duralumin are aluminium (95%), copper (4%), magnesium (0.5%) and manganese (0.5%).
(b) Solder are lead (50%) and tin (50%).
(c) Bronze are copper (80%), tin (18%) and zinc (2%).
(d) Invar are iron (63%), nickel (36%) and carbon (1%).

Miscellaneous Exercise

Solution 1.

(a) Bauxite: Aluminium is extracted from its main ore bauxite Al2O3.2H2O. It contains 60% Al2O3.
(b) Sodium hydroxide: Sodium hydroxide dissolves bauxite to form sodium meta aluminate, removes insoluble impurities from Al2O3 by forming red mud.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-1

Solution 2.

(a) Copper
(b) Iron
(c) Zinc
(d) Magnesium

Solution 3.

Arrangement of metal in decreasing order of reactivity are:
Sodium > Magnesium > Zinc > Iron > copper

Solution 4.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-4
(iii) In electrolytic process, the graphite acts as anode. The anode has to be replaced from time to time as it gets oxidized by the oxygen evolved at the anode.
(iv) The reaction that occurs at cathode is:
4Al3+ + 12e– → Al
(d) In construction the alloy of aluminium -duralumin is used because it is hard and resistant to corrosion.

Solution 5.

selina-icse-solutions-class-10-chemistry-metallurgy-mis-5
(d) In Aluminium thermite welding, the reduction with aluminium is highly exothermic and heat generated is sufficient to melt the metal.
Fe2O3 + 2Al → 2Fe + Al2O3 + Heat

Solution 1 (2005).

(i)

  1. Zinc: Froth Flotation, ZincBlende,Coke
  2. Aluminium: Bauxite,Cryolite, Sodium hydroxide solution

(ii)

  1. Sodium hydroxide.
  2. Cryolite

(iii) The formula of Cryolite is Na3AlF6.

Solution 1 (2006).

(a) Mercury
(b) Roasting
(c) CaSiO3
(d) Cryolite
(e) Graphite

Solution 1 (2007).

Acidic oxide(D)
Discharged at anode (F)
Covalent chlorides (I)
5,6,7 valence electrons (L)
Brittle(C)

Solution 1 (2008).

(i) A is made of carbon and B is thick graphite rod.
A → Cathode
B → Anode
(ii) Aluminium is formed at electrode A.
(iii) The two aluminium compound in the electrolyte C is Na3AlF6, Al2O3.
(iv) It is necessary to continuously replace electrode B from time to time as it gets oxidized by the oxygen evolved.

Solution 2 (2005).

(a) Stainless steel : Iron, Chromium
(b) Brass: Copper , Zinc

Solution 2 (2007).

selina-icse-solutions-class-10-chemistry-metallurgy-mis-2-2007

Solution 2 (2008).

Brass is an alloy of copper and Zinc.

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Selina Concise Chemistry Class 10 ICSE Solutions Study of Compounds – Study of Compounds – Ammonia

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Selina ICSE Solutions for Class 10 Chemistry Chapter 9 Study of Compounds – Ammonia

Exercise Intext 1

Solution 1.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-1

Solution 2.

The different forms of ammonia:

Gaseous ammonia(dry ammonia gas)
Liquid ammonia
Liquor ammonia fortis
Laboratory bench reagent

Solution 3.

Formula of liquid ammonia is: NH3.
Liquid ammonia is liquefied ammonia and is basic in nature. It dissolves in water to give ammonium hydroxide which ionizes to give hydroxyl ions.
selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-3

Solution 4.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-4

Solution 5.

(a) Lab preparation of ammonia:
2NH4Cl + Ca(OH)2 → CaCl2 +2H2O +2NH3
(b) The ammonia gas is dried by passing through a drying tower containing lumps of quicklime (CaO).
(c) Ammonia is highly soluble in water and therefore it cannot be collected over water.

Solution 6.

The drying agent used is CaO in case of ammonia.
Other drying agents like P2O5 and CaCl2 are not used. As ammonia being basic reacts with them.
6NH3 + P2O5 + 3H2O → 2(NH4)3PO4
CaCl2 +4NH3 → CaCl2.4NH3

Solution 7.

The substance A is Ammonium chloride and ‘B’ is Ammonia.
Reaction:
2NH4Cl + Ca(OH)2 → CaCl2 + 2H2O + 2NH3

Solution 8.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-8

Solution 9.

(a) Ammonium compounds being highly soluble in water do not occur as minerals.
(b) Ammonium nitrate is not used in the preparation of ammonia as it is explosive in nature and it decomposes forming nitrous oxide and water vapours.
(c) Conc. H2SO4 is not used to dry ammonia, as ammonia being basic reacts with them.
2NH3 + H2SO4 → (NH4)2SO4

Solution 10.

Preparation of Aqueous Ammonia: An aqueous solution of ammonia is prepared by dissolving ammonia in water. The rate of dissolution of ammonia in water is very high; therefore, back suction of water is possible. To avoid this, a funnel is attached to the outer end of the delivery tube with rubber tubing.

Procedure: Water is taken in a container and only a small portion of the mouth of funnel is dipped in water.

As ammonia dissolves in water at a higher rate than its production in the flask, the pressure in the funnel above water level decreases for a moment and water rushes into the funnel. As a result, the rim of the funnel loses its contact with water. Since, ammonia produced pushes the water down, the funnel comes in contact with water again. In this way, ammonia dissolves in water without back suction of water.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-ina-10

Exercise 1

Solution 1.

Physical properties of ammonia are:

Color: Colourless
Odour: Strong, Pungent chocking smell
Taste: Slightly bitter (alkaline ) taste
Physiological action: Non-Poisonous
Density: V.D = 8.5 Lighter than air
Nature: Alkaline
Liquefaction: easily liquefied at 10oC by compressing it at 6 atm. Pressure
Boiling Point: Liquid ammonia boils at -33.5oC
Freezing Point: Solid NH3 melts at -77.7oC
Solubility: Highly soluble in water, 1vol of water dissolves about 702 vols. of ammonia at 20oC and 1 atm. pressure.
Reaction:
2NH4Cl + Ca(OH)2 → CaCl2 + 2H2 + 2NH3

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-1

Solution 2.

Ammonia is less dense than air. By Fountain Experiment we demonstrate the high solubility of ammonia gas in water.
Balanced equation for the reaction between ammonia and sulphuric acid is:
2NH3 + H2SO4 → (NH4)2SO4

Solution 3.

(a) Ammonia is basic in nature.
(b) Copper oxide because CuO is less reactive can be reduced by C, CO or by hydrogen whereas Al2O3, Na2O, MgO are reduced by electrolysis.

Solution 4.

(a) The formula of the compound is Mg3N2.
(b) Balanced equation :
Mg3N2 + 6 H2O → 3Mg(OH)2 + 2 NH3
(c) Ammonia is a reducing agent and reduces less active metal oxide to its respective metal.

Solution 5.

Reducing property.

Solution 6.

When a piece of moist red litmus paper is placed in a gas jar of ammonia it turns blue.

Solution 7.

(a) The gas is ammonia.
(b) The formula is NH3.
(c) Uses of ammonia:

  1. It is used in the industrial preparation of nitric acid by Ostwald process.
  2. It is used in the manufacture of fertilizers such as ammonium sulphate, ammonium nitrate, ammonium phosphate.
  3. It is used in the manufacture sodium carbonate by Solvay process.
    NaCl + NH3 + CO2 + H2O → NaHCO3 + NH4Cl

Solution 8.

Equation:
CuSO4 + 2NH4OH → Cu(OH)2 ↓ + NH4]2SO4
pale blue

Ammonia solution in water gives a blue precipitate when it combines with a solution of copper salt.

The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper[II] sulphate, an azure blue(deep blue)soluble complex salt.
Cu(OH)2 +(NH4)2SO4 +2NH4OH → [Cu(NH3)4]SO4 + 4H2O

Solution 9.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-9

Solution 10.

(a) Liquid ammonia takes a lot of energy to vaporize .This heat is taken from the surrounding bodies which are consequently cooled down. Thus it is used as a refrigerant in ice plant.
(b) Ammonia emulsifies or dissolves fats, grease so it is used to remove grease from woolen clothes.
(c) Aqueous solution of ammonia gives pungent smell because of the presence of ammonia.
(d) Aqueous ammonia when dissolved in water breaks into ions which help in the conductance of electricity.
selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-10

Solution 11.

(a) AlN + 3H2O → Al(OH)3 +NH3
(b) 2NH3 + 3PbO → 3Pb + 3H2O + N2
(c) 8NH3 +3Cl2 → N2 + 6NH4Cl
(d) 2NH3 + CO2 → NH2CONH2 + H2O
(i) Ammonia act as reducing agent is explained by equation (c).
(ii) Urea the nitrogenous fertilizer is prepared from equation (d).

Solution 12.

(a) A Dirty green precipitate of Fe(OH)2 is obtained when ammonium hydroxide is added to ferrous sulphate.
(b) Liquid ammonia is liquefied ammonia.
(c) Finely divided Iron is used in Haber process.
(d) Quicklime is a drying agent for NH3.
(e) Ammonium salts when heated with caustic alkali.

Solution 13.

(a) Dirty green ppt. of Ferrous hydroxide is formed which is insoluble in excess of NH4OH.
FeSO4 + 2NH4OH → [NH4]2SO4 + Fe(OH)2

(b) Reddish brown ppt. of ferric hydroxide is formed which is insoluble in ammonium hydroxide.
FeCl3 + 3NH4OH → 3NH4Cl + Fe(OH)3

(c) White ppt. of lead hydroxide is formed which is insoluble in NH4OH.
Pb(NO3)2 + 2NH4OH → 2NH4NO3 + Pb(OH)2

(d) White gelatinous ppt. of Zinc hydroxide is formed which is soluble in NH4OH.
Zn(NO3)2 + 2NH4OH → 2NH4NO3 + Zn(OH)2

Solution 14.

When correct amount of ammonium hydroxide is added drop wise to solutions of the metallic salts, ppts. (coloured generally) are formed. They help us to identify their metal ions.
Two equations:
FeSO4 +2NH4OH → (NH4)2SO4 + Fe (OH)2
(Green) (Dirty green)
shows the presence of Fe+2 ion.
FeCl3 + 3NH4OH → 3NH4Cl + Fe (OH)3
(Brown) (Reddish brown)
shows the presence of Fe+3 ion.

Solution 15.


NH4Cl on strong heating sublimes to form dense white fumes which condense to white powdery mass on cooler parts of the tube whereas no white fumes on heating NaCl.

(b) When ammonium hydroxide is added drop wise to solution to be tested.
Ferrous salt gives dirty green ppt.
Ferric salt gives reddish brown ppt of their hydroxides.

(c) (NH4)2SO4 on warming with NaOH sol. gives NH3 gas. Sodium sulphate does not liberate NH3 gas.

Solution 16.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-16

Solution 17.

(a) In the presence of Platinum at 800oC, ammonia reacts with oxygen to give nitric oxide and water vapour.
Procedure:
Pass dry ammonia gas and oxygen gas through inlets over heated platinum placed in the combustion tube, which in the heated state emits reddish glow.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-17
Observation:
Reddish brown vapours of nitrogen dioxide are seen in the flask due to the oxidation of nitric oxide.
The platinum continues to glow even after the heating is discontinued since the catalytic oxidation of ammonia is exothermic.

(b) Two reactions to show reducing property of ammonia are:
8NH3 +3Cl2 → N2 + 6NH4Cl
2NH3 +3CuO → 3Cu + 3H2O +N2

Solution 18.

(i) Neutralization
NH3 + HCl → NH4Cl

(ii) Thermal dissociation
NH4Cl → NH3 + HCl

(iii) Ammonia
NH4Cl + NaOH → NH3 + NaCl + H2O

Solution 19.

(a) Ammonia
(b) Hydrogen chloride and chlorine gas.

(c) (i) Ammonium chloride
(i) Ammonium nitrate
(ii) Ammonium carbonate

(d) Acidic gas: HCl
Basic gas: Ammonia
Neutral gas: NH4Cl

(e) Silver chloride
(f) Nitrogen
(g) Magnesium nitride
(h) Lead oxide
(i) Ammonium chloride

Solution 20.

CuSO4 + 2NH4OH → (NH4)2SO4 + Cu(OH)2 [Pale blue]
The cation present in solution B is Copper (Cu+2).
The colour of solution B is Blue.
The pale blue precipitate of copper hydroxide dissolves in excess of ammonium hydroxide forming tetraamine copper[II] sulphate, an azure blue(deep blue) soluble complex salt.
Cu(OH)2 + (NH4)2SO4 +2NH4OH →  [Cu(NH3)4]SO4 + 4H2O

Solution 21.

Three ways in which ammonia gas can be identified is:

  1. It has a sharp characteristic odour.
  2. When a glass rod dipped in HCl is brought in contact with the gas white colour fumes of ammonium chloride are formed.
  3. It turns moist red litmus blue, moist turmeric paper brown and phenolphthalein solution pink.

Solution 22.

(a) Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3
(b) 2NH3 + 3CuO 3Cu + 3H2O + N2
Ammonia acts as a reducing agent. It reduces metallic oxide to give metals, water vapour and nitrogen.
(c) 8NH3 +3Cl2 N2 + 6NH4Cl
(d) 4 NH3 +5O2 6H2O + 4NO +Heat
Ostwald process starts with the catalytic oxidation of ammonia to manufacture nitric acid in the presence of catalyst platinum.

Solution 23.

As the ‘A’ turns red litmus blue it is a base. Now the gas ‘A’ combines with ‘B’ in presence of Catalyst to give colourless gas Nitrogen monoxide. It reacts with oxygen to give brown gas which is Nitrogen dioxide.
A= NH3
B= O2
C=NO
D=NO2

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-23

Solution 24.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-24

Solution 25.

(a) The main refrigerants used are Freon chlorofluorocarbons (CFC). They deplete ozone layer. The chlorofluorocarbons are decomposed by ultraviolet rays to highly reactive chlorine which is produced in the atomic form.selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-25
This causes depletion of ozone layer and chlorine monoxide so formed reacts with atomic oxygen and produces more chlorine free radicals.
ClO + O → Cl + O2
Again this free radical destroys ozone and the process continues thereby giving rise to ozone depletion.
(b) Liquid ammonia can be used as a refrigerant, as an alternative for chlorofluorocarbons.
(c) Advantages of ammonia as refrigerant:

  1. Ammonia is environmentally compatible. It does not deplete ozone layer and does not contribute towards global warming.
  2. It has superior thermodynamic qualities as a result ammonia refrigeration systems use less electricity.

Ammonia has a recognizable odour and so leaks are not likely to escape.

Solution 26.

Disadvantages of ammonia as a refrigerant are as follows:

  1. It is not compatible with copper, so it cannot be used in any system with copper pipes.
  2. It is poisonous in high concentration although it is easily detectable due to its peculiar smell and since it is less dense than air it goes up in the atmosphere not affecting the life too much on earth.

Solution 27.

(a) Explosive: ammonium nitrate
(b) Medicine: ammonium carbonate
(c) Fertilizers: ammonium sulphate
(d) Laboratory reagent: ammonia solution

Solution 28.

(a) Dry air free from carbon dioxide and dry ammonia from Habers process.
(b) The catalyst used in the process is Platinum.
(c) The oxidizing agent used in the process is oxygen.
(d) Ratio of ammonia and air is 1:10.
(e) Quartz is acid resistant and when packed in layers help in dissolving nitrogen dioxide uniformly in water.

Solution 29.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-29

Solution 30.

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-30

Solution 1 (2003).

(a) Mg3N2 +6H2O → 3Mg(OH)2 + 2NH3
(b) Ammonia gas is collected in inverted gas jars by the downward displacement of air.
(c) Ammonia is not collected over water because it is highly soluble in water.
(d) Quicklime is used as a drying agent for ammonia.

Solution 1 (2004).

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-1-2004

Solution 1 (2005).

(a) It is the basic nature of ammonia molecule.
(b) Hydroxyl ion (NH3 + H2O → NH4+ + OH–)
(c) The red litmus paper turns blue in the solution.

Solution 1 (2006).

Pb(NO3)2+ NH4OH → 2NH4NO3+Pb(OH)2
The chalky white ppt. of lead hydroxide is formed.

Solution 1 (2007).

(a) HCl gas is more dense [V.D.=18.25,V.D. of ammonia =8.5] and it is collected by the upward displacement of air.
(b) NH3 + HCl → NH4Cl

Solution 2 (2005).

selina-icse-solutions-class-10-chemistry-study-compounds-study-compounds-ammonia-2-2005

Solution 2 (2007).

Balanced equation:

(a) 2NH3 + 3CuO → 3Cu + 3H2O + N2
(b) 2NH3 + 3Cl2 → N2 + 6HCl

Solution 2 (2008).

Magnesium Nitride

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