Frank ICSE Solutions for Class 10 Chemistry – Mole Concept And Stoichiometry

Frank ICSE Solutions for Class 10 Chemistry – Mole Concept And Stoichiometry

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Solution 1:

  1. Gay-Lussac’s law: It states that ‘when gases react, they do so in volumes which bear a simple ratio to one another, and also to the volume of the gaseous product, provided all the volumes are measured at the same temperature and pressure’.
  2. Avogadro’s law : It states that ‘Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules’.

Solution 2:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 1

Solution 3:
When stating the volume of a gas, the pressure and temperature should also be given because the volume of a gas is highly susceptible to slight change in pressure and temperature of the gas.

Solution 4:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 2

Solution 5:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 3

Solution 6:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 4

Solution 7:

  1. Gram atom: “The quantity of the element which weighs equal to its gram atomic mass is called one gram atom of that element”.
    For example: The gram atomic mass of hydrogen is 1g. So, 1g of hydrogen is 1 gram atom of hydrogen.
  2. Gram mole: “A sample of substance with its mass equal to its gram molecular mass is called one gram molecule of this substance or one gram mole”.
    For example: Gram molecular mass of oxygen is 32 g. So One gram mole of oxygen is 32g.

Solution 8:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 5

Solution 9:
Empirical formula:“Empirical formula of a compound is the formula which gives the number of atoms of different elements present in one molecule of the compound, in the simplest numerical ratio”.
Molecular formula: “Molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of the compound”.

Solution 10:

  1. The empirical formula of C6H6 is: CH
  2. The empirical formula of C6H12O6 is: CH2O.
  3. The empirical formula of C2H2 is: CH
  4. The empirical formula of CH3COOH is: CH2O.

Solution 11:
Three pieces of information conveyed by the formula H2O is that:

  1. It shows that there are 2 hydrogen atoms and 1oxygen atoms present in H2O.
  2. The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.
  3. It represents one molecule of compound water.

Solution 12:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 6

Solution 13:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 7

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Solution 14:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 8

Solution 15:

  1. Na2SO4.10H2O.
  2. C6H12O6.

Solution 16:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 9

Solution 17:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 10

Solution 18:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 11

Solution 19:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 12

Solution 20:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 13

Solution 21:
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Solution 22:
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Solution 23:
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Solution 24:
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Solution 25:
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Solution 26:
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Solution 27:
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Solution 28:
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Solution 29:
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Solution 30:
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Solution 31:
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Solution 1996-1:
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Solution 1996-2:
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Solution 1996-3:
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Solution 1996-4:
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Solution 1997-1:
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Solution 1997-2:
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Solution 1997-3:
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Solution 1997-4:
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Solution 1998-1:
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Solution 1998-2:
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Solution 1999-1:
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Solution 1999-2:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 36

Solution 1999-3:
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Solution 2000-2:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 38

Solution 2000-1:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 39

Solution 2000-1:
Gay – Lussac proposed this law.

Solution 2001-2:
Molecular mass of ethane = 30
According to Gay-Lussac’s law:
2 vol. of C2H6 requires= 7 vol. of oxygen
Vol. of C2H6 = 2 vol. = 100 L
Vol. of oxygen required = 7 vol. =350 L

Solution 2001-3:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 40

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Solution 2001-4:
The term is vapour density.

Solution 2001-5:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 41

Solution 2001-6:
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Solution 2001-7:
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Solution 2002-1:
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Solution 2002-2:
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Solution 2002-3:
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Solution 2003-1:
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Solution 2004-1:

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Solution 2004-2:
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Solution 2005-1:
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Solution 2006-1:
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Solution 2006-2:
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Solution 2006-3:
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Solution 2006-4:
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Solution 2007-1:
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Solution 2007-2:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 56

Solution 2008-1:
The gas laws which relates the volume of a gas to the number of molecules of the gas is avogadro’s law

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Solution 2008-2:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 57

Solution 2008-3:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 58
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 59

Solution 2009-2:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 60

Solution 2009-3:
The correct statement is that equal volumes of all gases under identical conditions contain the same number of molecules.

Solution 2009-4:
Frank ICSE Solutions for Class 10 Chemistry - Mole Concept And Stoichiometry 61

Solution 2009-1:
The relative molecular mass of the gas is 10.

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