Frank Chapter 5 Mole Concept and Stoichiometry ICSE Solutions and MCQ Class 10 Chemistry

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

1. State:
(a) Gay-Lussac's law of combining volumes
(b) Avogadro's law

Answer 

(a) 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'.
(b) 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'.

2. (a) Define atomicity of a gas.
(b) Differentiate between 2H and H₂.

Answer 

(a) Atomicity of a gas : The number of atoms in a molecule of a gas is called its atomicity. 
For example : Monoatomic means a gas molecule containing one atom. Similarly diatomic corresponds to two atoms and triatomic corresponds to three atoms in a molecule of a gas. 
(b) 

2H	H2
2H corresponds to two atoms of Hydrogen element.	H2 corresponds to a hydrogen molecule which contains two atoms of hydrogen.

3. When stating the volume of a gas, the pressure and temperature should also be given. Why?

Answer 

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.

4. (a) The relative atomic mass of Cl atom is 35.5 a.m.u. Explain this statement.
(b) What is the value of Avogadro's number?
(c) What is the value of molar volume of a gas at STP?

Answer 

(a) The relative atomic mass of Cl atom is 35.5 a.m.u. because chlorine consists of a mixture of two isotopes of masses 35 and 37 in the ratio of 3:1.
The average of the isotopic masses is 35×3 + 37×4 = 35.5
(b) The value of Avogadro's number is 6.023×10²³.
(c) The value of molar volume of a gas at STP is 22.4 dm³ (litre) or 22400 cm³(ml).

Concept Insight:
(a) Isotopes are atoms of same having same atomic number but different mass number.
(c) One mole of any gaseous molecules occupy 22.4dm³ at standard temperature and pressure (STP). This volume is known as molar volume.

5. Define the terms:
(a) Vapour density.
(b) Molar volume.
(c) Relative atomic mass.
(d) Avogadro's number.
(e) Relative molecular mass.

Answer 

(a) Vanour density: It is Density of a gas, expressed as the mass of a given volume of the gas divided by the mass of an equal volume of a reference gas (such as hydrogen or air) at the same temperature and pressure.
(b) Molar volume: One mole of any gaseous molecules occupy 22.4dm³ at standard temperature and pressure (STP). This volume is known as molar volume.

"The molar volume of a gas can be defined as the volume occupied by one mole of a gas at standard temperature and pressure."

(c) Relative atomic mass: The relative atomic mass or atomic weight 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.
Relative atomic mass = Mass of 1 atom of the element/1/12 of the mass of one C¹² atom.

(d) Avogadro's number: Avogadro's number is defined as the number of atoms present in 12g of C¹² isotope i.e. 6.023 × 10²³ atoms.

• It is number of elementary units i.e. atoms, ions or molecules present in one mole of a substance.
• It is denoted by N_A.

(e) Relative molecular mass: "The relative atomic mass (or molecular weight) of an element or a compound is the number that represents how many times one molecule of the substance is heavier than 1/12 of the mass of an atom of carbon -12.

6. (a) What are the main applications of Avogadro's law?

(b) How does Avogadro's law explain Gay Lussac's law of combining volumes?

Answer 

(a) The main applications of Avogadro's law are:

• Explanation of Gay-Lussac's Law
• Determination of atomicity of gases
• Determination of the molecular formula of a gaseous compound.
• Establishes relationship between the relative vapour density of a gas and its relative molecular mass.
• Establishes the relationship between gram molecular weight and volume of a gas at STP.

(b) Exnlanation of Gay–Lussac’s  Law: Gay-Lussac had experimentally determined that one volume of hydrogen and one volume of chlorine react to produce two volumes of hydrogen chloride gas.
According to Avogadro's law, if :
1 volume of hydrogen contains n molecules of the gas then 1 volume of chlorine also contains n molecules of the gas. Therefore 2 volume of hydrogen chloride contain 2n molecules of the gas.

but hydrogen and chlorine are diatomic.
So, 2 atoms + 2 atoms → 2 molecules
1 atom + 1 atom → 1 molecule
i.e.  1 molecule of hydrogen chloride is formed when 1 atom of hydrogen combines with 1 atom of chlorine. Thus Avogadro's law explains Gay - Lussac's combining volumes. 

7. Explain the terms:
(a) Gram atom
(b) Gram mole
(c) Mole

Answer 

(a) 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.
(b) 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.

8. Calculate the relative molecular masses of (use K = 39, Cl = 35.5, O = 16, C = 12, H = 1, Na = 23, N = 14, S= 32)
(a) Potassium chlorate
(b) Sodium acetate
(c) Chloroform
(d) Ammonium sulphate

Answer 

(a) The relative molecular mass of Potassium chlorate (KClO₃) is :
[atomic mass of 1 K atom + Atomic mass of 1 Cl atom + atomic mass of 3O atoms]
= 39+ 35.5 +16×3 = 122.5
(b) The relative molecular macs of Sodium acetate (CH₃COONa) is:

[atomic mass of 2 C atom + Atomic mass of 3 H atom + atomic mass of 2 O atoms + atomic mass of 1 Na atom]
= 12×2 + 3×1 + 16×2 + 23 = 82
(c) The relative molecular mass of Chloroform (CH₃Cl):
[atomic mass of 1 C atom + Atomic mass of 3 H atom + atomic mass of 1 Cl]
= 12 + 3×1 + 35.5 = 50.5
(d) The relative molecular mass of Ammonium sulphate (NH₄)₂SO₄ is: 

[atomic mass of 2 N atom + Atomic mass of 8 H atom + atomic mass of 1 S + atomic mass of 4 O atom]
= 14×2 + 8×1 + 32 + 16×4 = 132

9. Explain the terms empirical formula and molecular formula.

Answer 

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.

10. Give the empirical formula of:
(a) C₆H₆
(b) C₆H₁₂O₆
(c) C₂H₂
(d) CH₃COOH

Answer 

(a) The empirical formula of C₆H₆ is: CH
(b) The empirical formula of C₆H₁₂O₆ is: CH₂O.
(c) The empirical formula of C₂H₂ is: CH
(d) The empirical formula of CH₃COOH is: CH₂O.

11. Give three pieces of information conveyed by the formula H₂O.

Answer 

Three pieces of information conveyed by the formula H₂O is that:
It shows that there are 2hydrogen atoms and 1oxygen atoms present in H₂O.
The hydrogen and oxygen atoms are present in simplest whole number ratio of 2:1.
It represents one molecule of compound water.

12. What do you understand by the term mole? How many elementary units are in one mole of a substance?

Answer 

"A mole is defined as the amount (mass) of a substance containing elementary particles like atoms, molecules or ions equal to Avogadro's number i.e. 6.023×10²³ .
Or a collection of 6.023 × 10²³ particles is called mole. 
Number of elementary units present in one mole of a substance is 6.023 × 10²³ . 

13. Fill in the blanks:
(a) Molecular weight of a gas is twice its ______.
(b) Mass of 22.4 litre of a gas at STP is ______.
(c) One gram atom of an element contains ______ atoms.
(d) One a.m.u. is the mass of ______ atom of C^₁₂.
(e) Avogadro's number is equal to ______.

Answer 

(a) Vapour density. 
(b) One mole of gas. 
(c) Quantity of element which weighs equal to its gram atomic mass 
(d) one 
(e) 6.023×10²³ 

14. Prove the following :
(a) Oxygen is a diatomic molecule. 
(b) 2 × V.D = Molecular mass.
(c) One mole of any gas contains the same number of molecules.

Answer 

(a) Since one Oxygen molecule contains 2 atoms of oxygen, so it is a diatomic molecule.
O + O → O₂
(b) The molecular mass of the given compound is determined experimentally by vapour density method also, in which the vapour density of the compound is determined. Vapour density is related to molecular mass as:
Molecular mass = 2×vapour density.
(c) Since by definition of a mole it is defined as the amount (mass) of a substance containing elementary particles like atoms, molecules or ions equal to Avogadro's number i.e. 6.023 x 10²³ so one mole of any gas contains the same number of molecules.

15. Give examples of compounds whose:
(a) Empirical formula is the same as the molecular formula.
(b) Empirical formula is different from the molecular formula.

Answer 

(a) Na₂SO₄.10H₂O.
(b) C₆H₁₂O₆.

16. Nitrogen and oxygen gas react as illustrated by the equation given below :
N₂  + O₂  ⟶ 2NO
Calculate the volume of each reacting gas required to produce 1400 cm³ of nitric oxide. 

Answer 

Given reaction is: 
N₂  + O₂ ⟶ 2NO 

According to Gay-Lussac's law in the above reaction l volume of nitrogen combines with 1 volume of oxygen to produce 2 volumes of nitric oxide. 

The volume of nitric oxide produced is = 1400cm³ .
Let the volumes of nitrogen and oxygen gases be = x
Then,

So,  x + x = 1400
2x = 1400
x= 1400/2 = 700cm³  
Hence the volumes of reacting gases i.e. nitrogen and oxygen is 700 cm³  each. 

17. Calculate the number of molecules in 12.8 g of sulphur dioxide gas. Take Avogadro's number as 6 × 10²³. [ S = 32 , O = 16 ]

Answer 

Number of molecules in 12.8 g of sulphur dioxide gas. 
Molecular mass of SO₂ = 64 a.m.u. 
So, 64 g = 1 mole
12.8 g = 12.8/64 = 0.2 mole 
Now 1 mole of SO₂  contains = 6 × 10²³ 
= 10.2 × 10²³ molecules. 

18. Calculate the weight of one molecule of oxygen. [Avogadro's number = 6 × 10²³, O = 16 ]

Answer 

Weight of 6×10²³ molecules of oxygen = 32 g 
Weight of 1 molecule of oxygen = 32/6 ×10²³  = 5.33 × 10²³ g

19. Calculate the percentage composition of oxygen in lead nitrate [Pb(NO₃)₂]. [Pb = 207, N= 14, O = 16]

Answer

Pb + (N)₂  + (O)₆  
207 + 2×14 + 6×16 = 331. 
So, the molecular mass of Pb(NO₃)₂  = 331. 
331 by weight of  Pb(NO₃)₂  contain 96 parts by weight of oxygen. 
100 parts will contain = 96 5 100/331 = 29%
So, the percentage composition of oxygen in lead nitrate is 29% . 

20. The usefulness of a fertilizer depends upon percentage of nitrogen present in it. Find which of the following is a better fertilizer:
(a) Ammonium nitrate [NH₄NO₃]
(b) Ammonium phosphate [(NH₄)₃PO₄] (N=14,H=1,O=16,P=31)

Answer 

(a) Percentage of nitrogen in Ammonium nitrate [NH₄NO₃]
(N)₂ + (H)₄ + (0)₃
14×2 + 1×4 + 3×16 = 80.
So, the molecular mass of NH₄NO₃ = 80.
80 by weight of NH₄NO₃ contain 28 parts by weight of nitrogen.
100 parts will contain = 28× 100 /80 = 35%
So, the percentage composition of nitrogen in Ammonium nitrate is 35%.

(b) Percentage of nitrogen in Ammonium phosphate [(NHL₄)₃PO₄].
(N)₃ + (H)₁₂ + P + (0)₄
14×3 + 1×12+ 31 + 16×4 = 149.
So, the molecular mass of NH₄NO₃ =149.
149 by weight of (NH₄)₃PO₄ contain 42 parts by weight of nitrogen.
100 parts will contain = 42× 100 /149 =28.18
So, the percentage of nitrogen in ammonium phosphate is 28.18 %. Since the percentage of nitrogen is more in Ammonium nitrate so it is a better fertilizer.

21. A compound of lead has following percentage composition, Pb = 90.66%, O = 9.34%. Calculate empirical formula of a compound. [Pb = 207, O = 16]

Answer 

Empirical formula of a compound:

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
Pb	207	90.66	90.66/207 = 0.44	0.44/0.44 = 1	1 × 3 = 3
O	16	9.34	9.34/16 = 0.58	0.58/0.44 = 1.32	1.32 × 3= 3.96 = 4

Since the mole ratio for oxygen is fractional so we multiply  the whole ratio by 3 to make it a whole number. 
So, the empirical formula of the compound is Pb₃O₄ .

22. Empirical formula of a compound is CH₂O. If its empirical formula is equal to its vapour density, calculate the molecular formula of the compound.

Answer 

Empirical formula of the compound is CH₂O. 
Empirical formula mass = Atomic mass of C + Atomic mass of H + Atomic mass of O 
= 12 + 2×1 + 16 = 30. 
Now as empirical formula is equal to the vapour density then ;
Molecular mass = 2× vapour density 
= 2×30 = 60 
n = Molecular mass/Empirical formula mass 
= 60/30 = 2 
Molecular formula = n × empirical formula 
= 2 × (CH₂O)
= C₂H₄O₂ 
The molecular formula of the compound is C₂H₄O₂ . 

23. Potassium nitrate on strong heating decomposes as under : 
2KNO₂ ⟶ 2KNO₂  + O₂ 
Calculate : 
(i) Weight of potassium nitrite formed. 
(ii) Weight of oxygen formed when 5.05 g of potassium nitrate decomposes completely. 
(K = 39, O = 16, N = 14)

Answer 

(i) From the equation:
Molecular weight of KNO₃  = (Atomic mass of K + Atomic mass of N + Atomic mass of 0) = (39 + 14 + 16 x 3) = 101

Molecular mass of of KNO₂ = (39 +14 +16×2) = 85

From the reaction:

2 moles of KNO₃ gives = 2 moles of KNO₂
So, 202 g of KNO₃ gives = 170 g of KNO₂
(ii) Given equation is: 2KNO₃ → 2KNO₂ + O₂
Molecular mass of KNO₃ is : (Atomic mass of K + Atomic mass of N + Atomic mass of 0) = (39 + 14 + 16×3) = 101
Molerular mass of of KNO₂ = (39 +14 +16 x 2) = 85
Now, decomposition of 101 g of KNO₃ yield = 16 g of O₂
So, decomposition of 5.05 g of KNO₃ will yield = 16 x 5.05 /101 = 0.8 g Hence, when 5.05 g of potassium nitrate decomposes completely 0.8 g of oxygen is formed.

24. (a)What is the volume occupied by the following gases at STP?
(i) 48 g of oxygen.
(ii) 16 g of sulphur dioxide.
(b) Determine the molecular mass of a gas if 5g of it occupy a volume of 4 L at STP.

Answer 

(a) (i) Volume occupied by 48 g of oxygen.
As 32g of oxygen at STP occupies volume of = 22.4 L
48 g of oxygen at STP occupies volume of = 22.4 × 48/32 = 33.6 L
Hence, 48 g of sulphur dioxide will occupy a volume of 33.6 L
(ii) Volume occupied by 16 g of sulphur dioxide. 
64 g of sulphur dioxide at STP occupies volume of = 22.4 L
16 g of sulphur dioxide at STP occupies volume of = 22.4 × 16/64 = 5.6 L
Hence, 16 g of sulphur dioxide will occupy a volume of 5.6 L
(b) 4L of a gas at STP has mass = 5 g
22.4 L of a gas at STP will has molecular mass = 5 × 22.4/4 = 28
So, the molecular mass of a gas will be 28. 

25. What weight of sulphuric acid will be required to dissolve 3 g of magnesium carbonate ?
[ Mg = 24, C =12, O = 16]
MgCO₃  + H₂SO₄  ⟶ MgSO₄ + H₂O + CO₂  

Answer 

Molecular formula of MgCO₃  is = 84 
Molecular formula of H₂SO₄  = 98 
Now if, 84 g of MgCO₃  requires = 98 g of H₂SO₄  
3 g of MgCO₃  will require = 98 × 3/84 = 3.5 g 
So  3.5 g of sulphuric acid will be required to dissolve 3 g of magnesium carbonate. 

26. Calculate the percentage of water in ferrous sulphate crystals [Fe = 56, S = 32, O =16, H = 1].

Answer 

Ferrous sulphate is FeSO₄ . 7H₂O 
Molecular mass of Ferrous sulphate is FeSO₄ . 7H₂O is :
Atomic mass of Fe + Atomic mass of S + Atomic mass of H + Atomic mass of O 
56 +32 +1×14 + 16×11 = 278 
278 parts by weight of crystals contain 126 parts of water 
100 parts will contain = 126 × 100/278 = 45.32 %
So, the percentage of water in ferrous sulphate crystals is 45.32 % 

27. An organic compound has the following percentage composition: C = 12.76%, H = 2.13%, Br = 85.11%. The vapour density of the compound is 94. Find out its molecular formula.

Answer 

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
C	12	12.76	12.76/12 = 1.06	1.06/1.06 = 1	1
H	1	2.13	2.13/1 = 2.13	2.13/1.06 =2	2
Br	80	85.11	85.11/80 = 1.06	1.06/1.06 = 1	1

So the Empirical formula of the compound will be CH₂Br. 
Now the Empirical formula mass will be = Atomic mass of C + Atomic mass of H + Atomic mass of Br 
= 12 + 1×2 + 80 = 94 
Now as Molecular mass = 2× vapour density 
= 2 × 94 = 188. 
So n = Molecular mass/ Empirical formula mass 
= 188/94 = 2 
Molecular formula of the compound is = n × empirical formula 
= 2 × ( CH₂Br)
= C₂H₄Br₂ .  

28. Two oxides of a metal (M) have 20.12% and 11.19% oxygen. The formula of the first oxide is MO. Determine the formula of the second oxide.

Answer 

Calculation of molar mass of M from first oxide : 
Let us assume the atomic mass of M as x. 
Atomic mass of Oxygen  = 16 

Element	Percentage	Relative number of moles	Simplest mole ratio
M	100 –20.12 = 79.88	79.88/x	79.88/1.25x
O	20.12	20.12/16 = 1.25	1.25/1.25 = 1

So, molar mass of M, x = 63.5 
Calculation of formula of second oxide : 

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
M	63.5	88.81	88.81/63.5 = 1.4	1.4/0.69 = 2	2
O	16	11.19	11.19/16 = 0.69	0.69/0.69 = 1	1

So formula of second oxide is M₂O  .

29. Calculate the volume of air at STP, required to convert 300 mL of sulphur dioxide to sulphur trioxide. Air contains 21% of oxygen by volume.

Answer 

Conversion of sulphur dioxide to sulphur trioxide follows the following balanced equation: 

According to Gay-Lussac's law :

Volume of SO₂ = 4 vol. = 300 mL
Volume of O₂ = 2 vol. = 300× 2/4 = 150 mL
Volume of oxygen required = 21% = 150 mL
Volume of air required at STP = 100% = 100 × 150/21 = 714.28 mL
So, the volume of air at STP ,  required to convert 300 mL of sulphur dioxide to sulphur trioxide is 714.28 mL . 

30. 10 g of NaCl solution is mixed with 17 g of silver nitrate solution. Calculate the weight silver chloride precipitated. 
AgNO₃  + NaCl → NaNO₃  

Answer 

Given reaction is :
AgNO₃  + NaCl ⟶ AgCl + NaNO₃  
Molecular mass of AgNO₃  is 170 
Molecular mass of AgCl is 143.5 
170 g of AgNO₃ produces = 143.5 g of AgCl
14 g of AgNO₃ will produce = 143.5× 17/170 = 14.35 g
So, the weight of silver chloride precipitated is 14.35 g 

31. A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10^-12 g. Calculate the number of carbon atoms in the signature.

Answer 

12 g of carbon contains = 6.023 × 10²³ number of  carbon atoms 
10^-12 g of carbon will contain = 6.023×10²³×10^-12/12
= 0.5019 × 10¹¹ 
= 5.019 × 10¹⁰ 
So, the number of carbon atoms in the signature is 5.019 × 10¹⁰.

32. Under the same conditions of temperature and pressure you collect 2 L of carbon dioxide, 3L of chlorine, 5 L of hydrogen, 4 L of nitrogen and 1 L of sulphur dioxide. In which gas sample will there be :
(a) The greatest number of molecules.
(b) The least number of molecules.
Justify your answer.

Answer 

(a) Calculation of number of molecules in each gas sample:
(i) 2 L of carbon dioxide:
22.4 L of carbon dioxide has = 6.023× 10²³ molecules
2 L of carbon dioxide will have = 6.023×10²³× 2 /22.4 = 0.5377×10²³ molecules

(ii) 3L of chlorine
22.4 L of chlorine has = 6.023×10²³ molecules
3 L of chlorine will have = 6.023×10²³× 3/22.4 = 0.8066×10²³ molecules

(iii) 51 of hydrogen
22.4 L of hydrogen has = 6.023 x 10²³ molecules
5 L of hydrogen will have = 6.023 x 10²³× 5/22.4 = 1.34×10²³ molecules
(iv) 4 L of nitrogen
22.4 L of nitrogen has = 6.023× 10²³ molecules
4 L of nitrogen will have = 6.023× 10²³× 4/22.4 = 1.07×10²³ molecules
(v) 1 L of sulphur dioxide
22.4 L of sulphur dioxide has = 6.023 × 10²³ molecules
1 L of sulphur dioxide will have = 6.023× 10²³× 1/22.4 = 0.27×10²³ molecules
From the above calculation of number of molecules in different gases we can conclude that the:
(a)The greatest number of molecules are present in 5 L of hydrogen gas sample.
(b) The least number of molecules is in 1 L of sulphur dioxide gas sample.

33. What volume of hydrogen sulphite at STP will burn in oxygen to yield 12.8 g sulphur dioxide according to the equation  ?
2H₂S + 3O₂ ⟶ 2H₂O + 2SO₂ 
(b) For the volume of hydrogen sulphide determined in (a) above, what volume of oxygen would be required for complete combustion ?  

Answer 

(a) 64 g of SO₂  will be produced at STP from = 22.4L of H₂S
12.8 g of SO₂  will be produced at STP from= 22.4 × 12.8 /64 = 4.48 L
So, 4.48 L of hydrogen sulphide at STP will burn in oxygen to yield 12.8g sulphur dioxide. 

(b) From the equation we know that 2 moles of H₂S burn in presence of 3 moles of Oxygen so:
44.8 L of H₂S requires = 67.2 L of oxygen
4.48 L of H₂S will require = 67.2/44.8× 4.48 = 6.72 L
So, 6.72 L of oxygen would be required for complete combustion. 

34. Find the total percentage of oxygen in magnesium nitrate crystal Mg(NO₃)₂.6H₂O.
[H = 1, N = 14, O = 16, Mg = 24]

Answer 

Molecular mass of Mg (NO₃)₂.6 H₂O is = 256 
Now, 256 parts by weight of crystal contains 192 parts by weight of oxygen. 
So 100 parts by weight will contain = 192× 100/256 = 75%
Hence, total percentage of oxygen in magnesium nitrate crystal Mg(NO₃)₂. 6 H₂O is 75 %  . 

35. A compound contains 87.5% by mass of nitrogen and 12.5% by mass of hydrogen. Determine the empirical formula of this compound.

Answer 

Empirical formula of the compound is as : 

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
N	14	87.5	87.5/14 =6025	6.25/6.25 = 1	1
H	1	12.5	12.5/1 = 12.5	12.5/6.25 = 2	2

So, the empirical formula of the compound is NH₂ . 

36. (a) Is it possible to change the temperature and pressure of a fixed mass of a gas without changing its volume? Explain your answer.
(b) Define or explain the meaning of term 'molar volume'.

Answer 

(a) No. it is not possible to change the temperature and pressure of a fixed mass of a gas without changing its volume because all the three vanables are interrelated to each other by the gas equation as:
PV/T = K (constant)
Hence if we change any one or two of the variables in the above equation then automatically third variable also has to change to make equation 1 equal to a constant. 

(b) "The molar volume of a gas can be defined as the volume occupied by one mole of a gas at standard temperature and pressure". It has been noticed that one mole of any gaseous molecules occupy 22.4 L of volume at standard temperature and pressure. 

37. What is the mass of nitrogen in 1000 Kg of urea [CO(NH₂)₂] ?
[H = 1, C= 12, N= 14, O = 16]

Answer 

Molecular mass of urea is = 60 
60 kg of urea has = 28 kg of nitrogen 
1000 kg or urea will have = 28× 1000/60 = 466.66 or 467 kg . 

38. (a) Calculate the empirical formula of the compound having 37.6% sodium, 23.1% silicon and 39.3% oxygen.(Answer correct to two decimal places) (O = 16, Na = 23, Si = 28)
(b) The empirical formula of a compound is C₂H₅. Its vapour density is 29. Determine the relative molecular mass of the compound and hence its molecular formula.

Answer 

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
Na	23	37.6	37.6/23 = 1.63	1.63/0.83 =1.9	2
Si	28	23.1	23.1/28 = 0.83	0.83/0.83 = 1	1
O	16	39.3	39.3/16 = 2.45	2.45/0.83 =2.9	3

So the empirical formula of the compound is Na₂SiO₃. (b) Given the empirical formula of the compound is C₂H₅.
Vapour density is = 29.
Empirical formula mass of the compound = 29

As, molecular mass = 2×vapour density = 2×29 = 58

So, molecular mass of the compound =58 Molecular formula = n× Empirical formula

Now, n = Molecular mass/Empirical formula mass = 58/29 =2

Molecular formula = 2× (C₂H₅) = C₄H₁₀
So, molecular formula of compound is = C₄H₁₀

39. (NH₄)₂Cr₂O₇ → N₂ +Cr₂O₃ + 4H₂O.
What volume of nitrogen at STP, will be evolved when 63 g of ammonium dichromate is decomposed ? (H = 1, N = 14, O = 16, Cr =52)

Answer 

Molecular mass of ammonium dichromate = 252 
Now, 252 g of ammonium dichromate evolves = 22.4 L of nitrogen at STP.
63 g of ammonium dichromate  will evolve = 22.4 × 63/252 = 5.6 L 
So, 63 g of ammonium dichromate will evolves 5.6 L of oxygen. 

40. (a) Calculate the percentage of boron (B) in borax (Na₂B₄O₇.10H₂O). [H = 1, B = 11, O = 16, Na = 23] , answer correct to 1 decimal place).
(b) (i) The compound A has the following percentage composition by mass:
C =26.7%, O = 71.1%, H = 2.2%. Determine the empirical formula of A.(Answer to one decimal place) (H=1,C=12,O=16)
(ii) If the relative molecular mass of A is 90, what is the molecular formula of A?

Answer 

(a) Molecular mass f borax Na₂B₄O₇ . 10H₂O = 382. 
382 parts by weight of borax contain 44 parts by weight of boron 
So 100 parts will contain = 44 × 100/382 = 11.5%
Percentage of boron (B) in borax ( Na₂B₄O₇ . 10H₂O) = 11.5%
(b) (i) 

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
C	12	26.7	26.7/12 = 2.2	2.2/2.2 = 1	1
O	16	71.1	71.1/16/ = 4.44	4.44/2.2 = 2	2
H	1	2.2	2.2/1 = 2.2	2.2/2.2 = 1	1

So the empirical formula of the compound is CO₂H . 
(ii) The relative molecule mass of A is 90 
Empirical Formula mass of compound is = 45 
Then, n = Molecular mass/Empirical Formula Mass 
= 90/45 = 2 
Molecular formula of compound = n × Empirical formula 
= 2×(CO₂H)
= C₂O₄H₂

41. Water can split into hydrogen and oxygen under suitable conditions. The equations representing the change is:

2H₂O (l) → 2H₂ (g) + O₂ (g)
(i) If a given experiment results in 2500cm³ of hydrogen being produced, what volume of oxygen is liberated at the same time under the same conditions of temperature and pressure?
(ii) Ammonia bums in oxygen and the combustion in the presence of a catalyst may be represented as:

2NH₃ (g) + 2½ O₂ (g) → 2NO (g) + 3H₂O (I)

What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?

Answer 

(i) Given equation is: 2H₂O → 2H₂ (g) + O₂ (g)
According to Gay-Lussac's law;
2 volume of water produces 2 volume of hydrogen and 1 volume of oxygen
i.e 2 volume of water produces = 2 volume of hydrogen = 2500 cm³
2 volume of water will produce = 1 volume of Oxygen = 2500/2 = 1250 cm³
i.e. = 1250 cm³
(ii) Given equation is: 2NH₃ (g) +2½ O₂ (g) → 2NO (g) + 3H₂O (l) Molecular mass of NO = 30
Molecular mass of H₂O = 18
From the equation:
2 moles of NO = 3 moles of H₂O
60 g of NO = 54 g of H₂O
1.5g of NO = 54× 1.5 /60 = 1.35 g of H₂O.

42. (a) Concentrated nitric acid oxidizes phosphorous to phosphoric acid according to the following equation:

P + 5HNO₃ → H₃PO₄ + H₂O + 5NO₂

(i) What mass of phosphoric acid can be prepared from 6.2 g of phosphorous?
(ii) What mass of nitric acid will be consumed at the same time?
(iii) What will be the volume of steam at the same time measured at 760 mm of Hg pressure and 273°C? [H = 1, N = 14, 0 = 16, P = 31]
(b) Ammonia may be oxidized to nitrogen monoxide in the presence of a catalyst according to the equation as:
4NH₃ + 5O₂ → 4NO + 6H₂0

If 27 L of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

Answer 

(a) Given equation is:
P + 5HNO₃ → H₃PO₄ + H₂O + 5NO₂
(i) Molecular mass of phosphorous = 31
Molecular mass of phosphoric acid = 98
31g of phosphorous produces = 98 g of phosphoric acid
6.2 g of phosphorous will produce = 98 × 6.2 /31 = 19.6 g
Hence,19.6 g of phosphoric acid can be prepared from 6.2 g of phosphorous.
(ii) Molecular mass of nitric acid = 63
31 g of phosphorous will consume= 63 g of nitric acid.
(iii) Moles of steam formed from 31g phosphorus = 1 mol
moles of steam from 6.2g phosphorus = l mol×6.2g/31g = 0.2 mol.
volume of steam produced at S.T.P = (0.2 mol) × (22.4 L/mol) = 4.48 litre.
Since the pressure (760 mm) remains constant, but the temperature (273 + 273) = 546 is doubled,
the volume of the steam also gets doubled
∴ volume of steam produced at 760 mm Hg and 273°C = 4.48 × 2 = 8.96 litres.

Volume of reactants = 4 vol. of ammonia  + 5 vol. of oxygen = 9 vol. 
9 vol. of reactants produces 4 vol. of Nitric oxide 
Therefore, 27 vol. of reactants will produce 4× 27 lit/9 = 12 litres of Nitric oxide 

43. If a crop of wheat removes 20 Kg of nitrogen per hectare of soil, what mass of the fertilizer calcium nitrate, Ca(NO₃)₂ would be required to replace nitrogen in 10 hectare field? (N = 14, O = 16, Ca = 40)

Answer 

Molecular weight of Ca(NO₃)₂ = 164
164 parts by weight of calcium nitrate contains 28 parts by weight of nitrogen.
28 Kg of nitrogen will be replaced by = 164 Kg of Ca(NO₃)₂  
20 Kg of nitrogen will be replaced by = 164 × 20 /28 = 117.14 kg
For, 1 hectare of field 20 Kg of nitrogen will be replaced by =117.14 Kg of Ca(NO₃)₂  
For, 10 hectare of field 20 Kg of nitrogen will be replaced by = 117.14 × 10 = 1171.4 Kg
Hence, 1171.4 Kg of the fertilizer calcium nitrate, Ca(NO₃)₂  would be required to replace nitrogen in 10 hectare field. 

44. (a) A vessel contains N molecules of oxygen at a certain temperature and pressure. How many molecules of sulphur dioxide can the vessel accommodate at the same temperature and pressure?
(b) Each of the two flasks contains 2.0 g of gas at the same temperature and pressure. One flask contains oxygen and the other hydrogen.
(i) Which sample contains the greater number of molecules?
(ii) If the hydrogen sample contains N molecules, how many molecules are present in oxygen sample?

Answer 

(a) As we know from Avogadro's law that under same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules so if a vessel contains N molecules of oxygen at a certain temperature and pressure and since the vessel remains same so same volume of sulphur dioxide will be present in the vessel and hence N molecules of sulphur dioxide can be accommodated in the vessel at the same temperature and pressure.
(b) (i) Flask having oxygen gas:

16 g of oxygen gas has = 6.023×10²³ molecules
2g of oxygen gas will have = 6.023×10²³/16×2 = 0.75×10²³
Flask having hydrogen:
1 g of hydrogen has= 6.023×10²³ molecules
2g of hydrogen gas will have = 6.023 × 10²³× 2 /1 = 12.05 × 10²³
So, hydrogen gas has greater number of molecules.
(ii) Amount of hydrogen and oxygen gases is same = 2 g
So, for oxygen 32 g of gas has =N molecules
Then, 2g of gas has = N/32 × 2 = 16
No of molecules of oxygen = N/16.

45. (a) If 112 cm³ of hydrogen sulphide is mixed with 120 cm³ of chlorine at STP, what is the mass of sulphur formed?
H₂S + Cl₂ → 2HCI + S

(b) Washing soda has the formula Na₂CO₃.10H₂0. What is the mass of anhydrous sodium carbonate left when all the water of crystallization is expelled by heating 57.2 g of washing soda?
(c) When excess lead nitrate solution was added to a solution of sodium sulphate, 15.1 g of lead sulphate was precipitated. What mass of sodium sulphate was present in the original solution?
Na₂SO₄ + Pb(NO₃)₂ → PbSO₄ + 2NaNO₃
(H = 1, C =12, 0 = 16, Na = 23, S = 32, Pb = 207)

Answer 

(a) At STP, 22400 cm³ of each H₂S and Cl₂ will give 32 g of sulphur i.e. 44800 cm³ of H₂S + Cl₂ gives = 32 g of S
(112 + 120) = 232 cm³ of H₂S + Cl₂ will give = 32 × 232 /44800 = 0.16 g of S

molecular weight of washing soda is 286.14 g
molecular weight of sodium carbonate is 106 g
286.14 g of Na₂CO₃.10H₂0 forms 106 g of sodium carbonate on heating 57.2 g of Na₂CO₃.10H₂0 forms = 106× 57.2 / 286.14 = 21.2 g
(c) Na₂SO₄ + Pb(NO₃)₂ → PbSO₄ + 2NaNO₂
molecular weight of Na₂SO₄ is 142 g
molecular weight of PbSO₄ is 303 g
303 g of PbSO₄ is formed by 142 g of Na₂SO₄
15.1 o of PbSO₄ Is formed by = 147×15.1/303 = 7.1 g of Na₂SO₄

46. Determine the empirical formula of the compound whose composition by mass is 42% nitrogen, 48% oxygen and 9% hydrogen.[N=14,O=16,H=1]

Answer 

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
N	14	42	42/14 =3	3/3 = 1	1
O	16	48	48/16= 3	3/3 = 1	1
H	1	9	9/1 = 9	9/3 = 3	3

So the empirical formula of the compound is NH₂OH.

47. When gases react their volumes bear a simple ratio to each other, under the same conditions of temperature and pressure. Who proposed this gas law?

Answer 

Gay-Lussac proposed this law.

48. What volume of oxygen would be required for complete combustion of 100 L of ethane  according to the following equation ?

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

Answer 

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

49. Chlorine, nitrogen, ammonia and sulphur dioxide gases are collected under the same conditions of temperature and pressure. 
Copy the following table which gives the volumes of the gases collected, and the number of molecules (X) in 20 L of nitrogen. You are to complete the table by giving the number of molecules in the other gases, in terms of X.

Gas	Volume (litres)	Number of molecules
Chlorine	10
Nitrogen	20	X
Ammonia	20
Sulphur dioxide	5

Answer 

(a) If 20 L of nitrogen has = X number of molecules 
Then, 10 L of chlorine will have = X  × 10/20 = X/2.
(b) If 20 L of nitrogen has  = X number of molecules
Then, 20 L of ammonia will have = X × 20/20 = X.
(c) If 20 L of nitrogen has = X number of molecules 
Then, 5 L of sulphur dioxide will have  = X × 5/20 = X/4.

Gas	Volume (litres)	Number of molecules
Chlorine	10	X/2
Nitrogen	20	X
Ammonia	20	X
Sulphur dioxide	5	X/4

50. Mention the term defined by the following sentence:
The mass of a given volume of gas compared to the mass of an equal volume of hydrogen.

 Answer 

The term is vapour density.

51. The reaction  4N₂O + CH₄ → CO₂ + 2H₂O + 4N₂   takes place in the gaseous state. If the volumes of all the gases are measured at the same temperature, and pressure, calculate the volume of dinitrogen oxide (N₂O), required to give 150 cm³ of steam. 

Answer 

According to Gay - Lussac's law: In the equation 

4N₂O + CH₄ → CO₂ + 2H₂O + 4N₂  
Vol. of  H₂O produced is = 2 vol. = 150 cm³ 
Vol. of N₂O required is  = 4 vol. = 150 × 4/2 = 300 cm³ 
300 cm³ of dinitrogen oxide (N₂O) is required to give 150cm³  of steam.

52. Calculate the percentage of phosphorous in the fertilizer superphosphate, Ca(H₂PO₄)₂. [Ca = 40, H =1, P =31, O = 16] (Correct to 1 decimal place)

Answer 

Molecular mass of fertilizer superphosphate, Ca(H₂PO₄)₂. = 234 
234 parts by weight of fertilizer contains 62 parts by weight of phosphorous 
So, 100 parts will contain = 62 × 100/234 = 26.5%

53. A metal M, forms a volatile chloride containing 65.5% Chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride. [M = 56, Cl = 35.5]

Answer 

Given, density of chloride relative to hydrogen = 162.5 
Percentage of chlorine = 65.5 %
Percentage of Metal M = 100 - 65.5 = 34.5%

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
M	56	34.5	34.5/56 = 0.62	0.62/0.62 = 1	1
Cl	35.5	65.5	65.5/35.5 = 1.85	1.85/0.62 = 2.98	3

54. Samples of the gases O₂, N₂, CO₂ and CO under the same conditions of temperature and pressure contain the same number of molecules represented by X. The molecules of Oxygen, occupy V litres and have a mass of 8 g. Under the same conditions of temperature and pressure:
(a) What is the volume occupied by:
(i) X molecules of N₂,
(ii) 3X molecules of CO? [C=12,N=14,O=16]
(b) What is the mass of CO₂ in g?
(c) In answering the above questions, which law have you used?

Answer 

(a) X molecules of N₂ occupies litres. 
3x molecules of CO occupies 3V litres. 
(b) X molecules of O₂ = 8/32 = 1/4 mole of O₂ 
So, X molecules of CO₂ = 1/4 molecule of CO₂ 
So, mass of CO₂  present in the sample = 1/4 × gram molecular mass of CO₂ 
= 1/4 × 44 = 11g
(c) Avogadro's law.  

55. Calculate the percentage of platinum in ammonium chloroplatinate (NH₄)₂PtCl₆. [N = 14, H = 1, Pt = 195, Cl =35.5] (Give your answer correct to the nearest whole number)

Answer 

Molecular formula of ammonium chloroplatinate (NH₄)₂PtCl₆ : 
2×(atomic mass of N + 8×atomic mass of H) + atomic mass of platinum + 6×atomic mass of chlorine 
2× (14+8) + 195 + 6×35.3 = 444
444 parts of ammonium chloroplatinate contains 195 parts by weight of platinum 
So, 100 parts will contain = 195 × 100/444 = 43.9% = 44%

56. The percentage composition of sodium phosphate, as determined by analysis is: 42.1% Na, 18.9% P, 39% of O. Find the empirical formula of the compound. [H =1, N =14, Na = 23, P = 31, Cl = 35.5, Pt = 195]

Answer 

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
Na	23	42.1	42.1/23 = 1.8	1.8/0.6 = 3	3
P	31	18.9	18.9/31 = 0.6	0.6/0.6 = 1	1
O	16	39	39/16 = 2.4	2.4/0.6 = 4	4

Empirical formula = Na₃PO₄

57. (a) What volume of oxygen is required to burn completely a mixture of 22.4 dm³ of methane and 11.2 dm³  of hydrogen into carbon dioxide and steam ? Equations of the reactions are given below : 
CH₄ + 2O₂ ⟶ CO₂ + 2H₂O
2H₂ + O₂ → 2H₂O
(b) The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular mass. Given 8 g of each gas at STP, which gas will contain the least number of molecules and which gas the most ? 

Answer 

One volume of methane requires oxygen = 2 vol.
So, Vol. of oxygen used = 2 × 22.4 = 44.8 dm³ 
2H₂  + O₂ ⟶ 2H₂ 0
2vol.   1vol.    2vol.
2 volume of hydrogen needs one volume of oxygen
So, Volume of oxygen used =22.4/2 = 11.2 dm³ 
Total volume of oxygen used = 44.8 + 11.2 = 56.0 dm³ 
(b) Calculation of number of molecules in each gas sample:
(i) 8 g of hydrogen:
1g of hydrogen = 6.023× 107²³  molecules
8g of hydrogen will have = 6.023×10²³ ×8 = 48.184 ×10²³  molecules
(ii) 8 g of oxygen:
32 g of oxygen has = 6.023× 10²³  molecules
8g of oxygen will have = 6.023 × 10²³/32 × 8 = 1.50×10²³ molecules
(iii) 8 g of carbon dioxide:
44 g of carbon dioxide has = 6.023× 10²³ molecules
8 g of carbon dioxide will have = 6.023×10²³/44 ×8 = 1.0×10²³ molecules
(iv) 8 g of sulphur dioxide:
64 of sulphur dioxide has = 6.023× 10²³ molecules
8 g of sulphur dioxide will have = 6.023×10²³/64×8 = 0.75 ×10²³   molecules
(v) 8 g of chlorine:
35.5 g of chlorine has = 6.023× 10²³ molecules
8 g of chlorine will have = 6.023 × 10²³ /71 × 8 = 0.68 × 10²³   molecules
Hence, chlorine gas will have least number of molecules. Hydrogen gas will have the most number of molecules.

58 . (a) A flask contains 3.2 g of sulphur dioxide. Calculate the following : 
(i) The moles of Sulphur dioxide present in the flask. 
(ii) The number of molecules of sulphur dioxide present in the flask. 
(iii) The volume occupied by 3.2 g of sulphur dioxide at STP. 
(b) The reaction of potassium permanganate (VII) with acidified iron (II) sulphate is given below :
2KmnO₄ + 10FeSO₄ + 8H₂O → K₂SO₄ + 2MnSO₄ + 5 Fe₂(SO₄)₃ + 8H₂O
If 15.8 g of potassium permanganate (VII) was used in the above reaction.

Answer 

(a) (i) Moles = Weight of substance in grams/ molecular weight
So, Moles of SO₂ = 3.2/64 = 0.05
(ii) Number of molecules = Moles ×6.023×10²³
So, number of molecules of SO. = 0.05 ×6.023×10²³
= 0.30115 × 10²³ = 0.302 × 10²³
(iii) 64 g of SO₂ occupy volume = 22.4L
So, 3.2 g of SO₂ will occupy a volume = 22.4× 3.2/64 = 1.12L

(b) Molecular weight of KmnO₄ = 39 + 55 + 16×4 = 158
Molecular weight of K₂SO₄ = 2×39 + 32 + 16×4 = 174
Molecular weight of FeSO₄ = 56 + 32 + 64 = 152
2 x 158 g of KmnO₄ yields = 174g of K₂SO₄
So, 15.8 g of KMnO₄ will yield = 174× 15.8/2 ×158 = 8.7g of K₂SO₄
174g of K₂SO₄ yields 152 g of FeSO₄
So, 8.7 g of K₂SO₄ will yield = 152× 8.7/174 = 7.6 g of FeSO₄
Hence, 7.6 g of iron (II) sulphate is used in the above reaction.

59. When heated, potassium permanganate decomposes according to the following equation :
2KmnO₄ → K₂MnO₄ + MnO₂ + O₂  
(a) Some potassium permanganate was heated in test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g , calculate the relative molecular mass of oxygen. 
(b) Given that the molecular mass of potassium permanganate is 158 g, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate ? (Molar volume at room temperature is 24 litres). [ K = 39, Mn = 55, O = 16]

Answer 

(a) Mass of one litre of oxygen gas liberated at room temperature = 1.32 g
Mass of one litre of hydrogen under the same conditions of temperature and pressure = 0.0825 g
Relative Molecular mass of oxygen = Weight of n molecule of O₂/Weight of 1/2 molecule of hydrogen
= 1.32× 2/0.0825 = 32 a.m.u.
(b) 2 KMnO₄ → K₂MnO₄ + MnO₂ + O₂
Molecular mass of KMnO₄ = 158
Molar volume of 02 at room temperature = 24 L
2× 158 g of KMnO₄ at room temperature yields = 24 L of O₂
15.8g of KmnO₄ will yield = 24× 15.8 /2 × 158 = 1.2 L of O₂ .

60. The equations given below relate to the manufacture of sodium carbonate
(Molecular weight of Na₂CO₃ = 106).
NaCl + NH₃ + CO₂ + H₂O → NaHCO₃ + NH₄Cl
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
Questions (a) and (b) are based on the production of 21.2 g of sodium carbonate.
(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate?
(b) To produce the mass of sodium hydrogen carbonate what volume of carbon dioxide, measured at STP would be required?

Answer 

(a) 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
Molecular mass of NaHCO₃ = 84
Molecular mass of Na₂CO₃ = 106
From the above reaction:
1 Na₂CO₃ is obtained from = 2 NaHCO₃
106 g of Na₂CO₃ is obtained from = 168 g of NaHCO₃
So, 21.2 g of Na₂CO₃ will be obtained from = 168× 21.2 /106 = 33.6 g of NaHCO₃
(b) NaCl + NH₃ + CO₂ + H₂O → NaHCO₃ + NH₄Cl
From the equation, 1 mole of CO₂  i.e. 22.4 L of CO₂ is used to produce = 1 mole of NaHCO₃  
Now as, 84 g of NaHCO₃  requires = 22.4 L of CO₂
33.6 g of NaHCO₃ will require = 22.4 × 33.6/84 = 8.96 L of CO₂

61. (a) The volumes of gases A, B, C and D are in the ratio, 1:2:2:4 under the same conditions of temperature and pressure.
(i) Which sample of gas contains the maximum number of molecules?
(ii) If the temperature and the pressure of gas A are kept constant then what will happen to the volume of A when the number of molecules is doubled?
(iii) If this ratio of gas volumes refers to the reactants and products of a reaction, which gas law is being observed?
(iv) If the volume of A is actually 5.6 dm3 at STP, calculate the number of molecules in the actual volume of D at STP.
(v) Using your answer from (iv), state, the mass of D if the gas is dinitrogen oxide (N₂O)
(b) Calculate the percentage of nitrogen in aluminium nitride. [Al = 27, N = 14]

Answer 

(a) (i) Gas D contains the maximum number of molecules. 
(ii) If the temperature and the pressure of gas A are kept constant, then the volume of the gas will get doubled. 
(iii) Gay - Lussac's law of combining volumes.
 
(v) 6 × 10²³ molecules is Avogadro's number of molecules contained in one gram mole of the substance if gas D is N₂O then, 
1 gram mole of N₂O = 2 × 14 +16 = 44 g
(b) Molecular mass of aluminium nitride (AlN₃) = 24 + 14 ×3 = 69 
Now, 69 parts by weight of aluminium nitride contains = 42 parts by weight of nitrogen 
So, 100 parts will contain = 42 × 100/69 = 60.86%
Hence, the percentage of nitrogen in aluminium nitride is 60.86% . 

62. (a) Determine the empirical formula of a compound containing 47.9% K, 5.5% beryllium and 46.6% fluorine by mass. [Be = 9, F = 19, K = 39]
(b) Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide ? The equation for the reaction is : 
3CuO + 2NH₃ ⟶ 3Cu + 3H₂O + N₂ 

Answer 

(a) Determination of empirical formula : 

Element	Atomic mass	Percentage	Relative Number of moles	Simplest mole ratio	Whole number ratio
K	39	47.9	47.9/39 = 1.2	1.2/0.6 = 2	2
Be	9	5.5	5.5/9 = 0.6	0.6/0.6 = 1	1
F	19	46.6	46.6/19 = 2.4	2.4/0.6 = 4	4

The empirical formula of compound = K₂BeF₄ 
(b) 3CuO + 2NH₃ ⟶ 3Cu + 3H₂O + N₂ 
Molecular mass of CuO = 80 
Volume occupied by 1 mole of NH₃ = 22.4 L 
From the equation : 
3 moles of CuO is reduced by 2 moles of ammonia 
For 240 g of CuO, volume of NH₃ consumed = 44.8 L 
For 120 g of CuO, volume of NH₃ consumed = 44.8× 120/240 = 22.4 L 

63. (a) Calculate the number of moles and number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?
(b) What is the vapour density of ethylene? (Avogadro's number = 6x10²³; Atomic weight of C = 12, H = 1; Molar volume = 22.4 litres at STP)

Answer 

(a) Molecular mass of ethylene (CH2-CH2) = 28 g
Number of moles = Given weight/ Molecular weight
= 1.4/28 = 0.05 moles
Now, number of molecules in 1 mole = 6.023× 10²³
So, number of molecules in 0.05 moles = 6.023× 10²³ × 0.05 = 0.3 × 10²³ = 3× 10²² molecules.
Volume occupied by 1 mole of ethylene = 22.4 L
So, volume occupied by 0.05 moles of ethylene = 22.4 × 0.05 = 1.12 L
(b) Vapour density = Molecular weight/2 = 28/2 = 14.

64. (a) Calculate the percentage of sodium in sodium aluminium fluoride (Na₃AIF₆). [F =19, Na = 23, Al = 27]
(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:
2CO + O₂ → 2CO₂ 
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

Answer 

(a) Molecular weight of sodium aluminium fluoride (Na₃AIF₆) = 210
Now, 210 parts by weight of Na₃AIF₆ contains = 69 parts by weight of sodium
So, 100 parts will contain = 69 × 100/210 = 32.8 or 33%
(b) 2CO + O₂ → 2CO₂
From the reaction:
2 volumes of CO consumes = 1 volume of O₂
So, 560 mL of CO consumes = 1/2 × 560 = 280 mL
Now, 2 volume of CO gives = 2 volume of CO₂
So, 560 mL of CO will give = 2 × 560 /2 = 560 ml

65. A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP)
NH₄NO₃ → N₂O + 2H₂O
(i) What volume of dinitrogen oxide is produced at the same time as 8.96 L of steam ? 
(ii) What mass of ammonium nitrate should be heated to produce 8.96 L of steam ? 
(iii) Determine the percentage of oxygen in the ammonium nitrate ? 

Answer 

(i) NH₄NO₃ → N₂O + 2H₂O
From the equation:
1 mole of NH₄NO₃ yields 2 mole of H₂O
So, 44.8 L of steam = 22.4 L of N₂O at STP
8.96 L of steam = 22.4× 8.96/44.8 = 4.48 L of N₂O at STP.

(ii) Molecular mass of NH₄NO₃ = 80
44.8 L of steam is liberated by = 80 g of NH₄NO₃
8.96 L of steam will be liberated by = 80× 8.96/44.8 = 16 L

(iii) 80 parts by weight of NH₄NO₃ contains 48 parts by weight of oxygen.
So, 100 parts will contain = 48× 100/80 = 60 %

66. A compound 'X' consists of 4.8% of C and 95.2% of Br by mass.
(i) Determine the empirical formula of this compound. [C = 12, Br = 80]
(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?
(iii) Name the type of chemical reaction by which X can be prepared from ethane.

Answer 

(i) Empirical formula of compound :

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
C	12	4.8	4.8/12 = 0.4	0.4/0.4 = 1	1
Br	80	95.2	95.2/80 = 1.19	1.19/0.4 = 3	3

Empirical formula of the compound is CBr₃ . 
(ii) Vapour density = 252
Empirical formula mass = 12 + 3×80 = 252
Molecular mass = 2 × Vapour density 
= 2 × 252  = 504 
Now, n = Molecular mass/Empirical Formula Mass 
= 504/252 = 2 
Molecular formula  = n × Empirical Formula 
= 2× (CBr₃ )
= C₂Br₆ .
(iii) This substance can be prepared by substitution method. 

67. Choose the correct alternative:

The gas laws which relates the volume of a gas to the number of molecules of the gas is:
(a) Avogadro's law
(b) Gay-Lussac's law
(c) Boyle's law
(d) Charle's law
Answer
(a) Avogadro's law
The gas laws which relates the volume of a gas to the number of molecules of the gas is Avogadro's law

68. The equation for the burning of octane is:
2C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
(i) How many moles of carbon dioxide are produced when one mole of octane burns ? 
(ii) What volume, at STP, is occupied by the number of moles determined in (i) ?
(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane ?
(iv) What is the empirical formula of octane ? 

Answer 

(i) 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
From the equation:
2 moles of C₈H₁₈ produces = 16 moles of CO₂
1 mole of C₈H₁₈ will produce = 16/2 = 8 moles of CO₂
So, 8 moles of CO₂ is produced.
(ii) Now, 1 mole of CO₂ occupies = 22.4 L at STP
So, 8 moles of CO₂ will occupy = 22.4 × 8 = 179.2 L at STP
(iii) Since 2 moles of C₈H₁₈ produces = 16 moles of CO₂ = 16 x 44 = 704 So by burning 2 moles of octane 704 g of CO₂ is produced.
(iv) Molecular formula of octane = C₈H₁₈
Its empirical formula will be = C₄H₉.

69. (a) (i) A compound has the following percentage coposition by mass: C is 14.4%,  H is 1.2% and Cl is 84.5%. Determine the empirical formula of this compound. 
(ii) The relative molecular mass of this compound is 168, so what is its molecular formula ? 
(iii) By what type of reaction could this compound be obtained from ethylene ?
(b) From the equation :
C + 2H₂SO₄ → CO₂ + 2H₂O + 2SO₂ 
Calculate :
(i) The mass of carbon oxidized by 49 g of sulphuric acid (C = 12, relative molecular mass of sulphuric acid = 98)
(ii) The volume of sulphur dioxide measured at STP, liberated at the same time. 

Answer 

(a) (i)

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
C	12	14.4	14.4/12 = 1.2	1.2/1.2 = 1	1
H	1	1.2	1.2/1 = 1.2	1.2/1.2 = 1	1
Cl	35.5	84.5	84.5/35.5 = 2.4	2.4/1.2 = 2	2

Empirical formula of the compound is CHCl₂ .
(ii) Now, empirical formula mass = 12 +1 + 35.5 × 2
= 12 + 1 + 71 = 84 
n = Relative molecular mass/Empirical Formula mass
= 168/84 = 2
So, molecular formula = n × (Empirical Formula)
= 2 × CHCl₂ .
= C₂H₂Cl₄ 
(iii) Addition reaction with chlorine. 

(b) (i) C + 2H₂SO₄ → CO₂ + 2H₂O + 2SO₂ 
From the equation :
2 moles of sulphuric acid oxidizes 1 mole of carbon 
i.e. 2× 98 g of sulphurice acid oxidizes 12 g of carbon 
So, 49 g of sulphuric acid will oxidize = 12 × 49/196 = 3g 
3g of carbon is oxidized by 46 g of sulphuric acid. 
(ii) Again from the equation : 
2 moles of sulphuric acid librates 2 moles of sulphur dioxide.
i.e. 196 g of sulphuric acid liberates = 44.8 dm³ of sulphur dioxide
49 g of sulphuric acid will liberate = 44.8 × 49/196 = 11.2 dm³ 
Hence, 11.2 dm³ of sulphuric dioxide is liberated at the same time.

70. Choose the correct answer:

A gas cylinder of capacity of 20 dm³ is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is
(a) 5
(b) 10
(c) 15
(d) 20
Answer
(b) 10
The relative molecular mass of the gas is 10.

71. (a) Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with the moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm; of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below:
2C₂H₂ (g) + 5O₂ (g) → 4CO₂ (g) + 2H₂O (g)
(b) A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 37.

Answer

(a) Applying Gay - Lussac's law on the equation : 

As 2 volume of acetylene requires = 5 volume of oxygen
So, 200 cm³ of acetylene will require = 5 × 200/2 = 500 cm³
Now further, 2 volume of acetylene produces = 4 volume of carbon dioxide
So, 200 cm³ of acetylene will produce = 4 × 200/2 = 400 cm³
Hence, 500 cm³ of oxygen and 400 cm³ of carbon dioxide Ls formed.

(b) Calculation of empirical formula:
Percentage of hydrogen = 12.5%
Percentage of nitrogen = 100 -12.5 = 87.5%

Element	Atomic mass	Percentage	Relative number of moles	Simplest mole ratio	Whole number ratio
N	14	87.5	87.5/14 = 6.25	6.25/6.25 = 1	1
H	1	12.5	12.5/1 = 12.5	12.5/6.25 = 2	2

Empirical formula = NH₂ 

Given molecular mass = 37 
Empirical formula mass = 16 
n = Molecular mass/Empirical Formula mass
= 37/16 = 2.3 or approximately 2 
Molecular formula = n × Empirical formula 
= 2 × NH₂ 
= N₂H₄ 

72. Correct the following statement:
Equal masses of all gases under identical conditions contain the same number of molecules.

Answer 

The correct statement is that equal volumes of all gases under identical conditions contain the same number of molecules.

73. (a) A gas cylinder contains 24 × 10²⁴ molecules of nitrogen gas. If Avogadro's number is 6 × 10²³ and the relative atomic mass of nitrogen is 14, calculate:
(i) Mass of nitrogen gas in the cylinder.
(ii) Volume of nitrogen at SW in dm³ .
(b) Commercial sodium hydroxide weighing 30 g has some sodium chloride in it. The mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is:
NaCl + AgNO₃ → AgCl + NaNO₃
[Relative molecular mass = 58, AgCl = 143]
(c) A certain gas 'X' occupies a volume of 100 cm³ at STP and weighs 0.5 g. Find its relative molecular mass.

Answer 

(a) (i) Molecular weight of nitrogen= 28
As 6.023×10²³ molecules of nitrogen weigh =28
24×10²³ molecules will weigh = 28×24× 10²³/6.023 x 10²³
= 28×40 = 1120g
(ii) As 6.023×10²³ molecules of nitrogen occupies = 22.4 dm³ at STP
24 x 10²³ molecules will occupy = 22.4×24× 10²³/6.023×10²³ = 896 dm³ 

(b) NaCl + AgNO₃ → AgCl + NaNO₃
As 143 g of AgCl is obtained from =58 g of NaCl
So, 14.3 g of AgCl will be obtained from = 58× 14.3/143 = 5.8 g of NaCl Weight of commercial NaOH = 30 g
Percentage of NaCl in NaOH = 5.8× 100/30 = 19.33%
(c) As at STP 100 cm³ of gas weighs = 0.5 g
So, at STP 22400 cm³ of gas will weight = 0.5× 22400 /100 = 112 g

74. (i) LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reaction can be presented as :

(ii) Calculate the percentage of nitrogen and oxygen in ammonium nitrate [Relative molecular mass of ammonium nitrate is 80, H=1, N=14, O = 16].

Answer 

(ii) Molecular mass of NH₄(NO₃) = 80
H = 1, N = 14, O = 16
% of nitrogen
As 80 g of NH₄ (NO₃) contains 28 g of nitrogen,

75. 4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.
(i) Write the equation for the reaction.
(ii) What is the mass of 4.5 moles of calcium carbonate? Relative molecular mass of calcium carbonate is 100)
(iii) What is the volume of carbon dioxide liberated at STP?
(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111)
(v) How many moles of HCl are used in this reaction

Answer 

(i) Equation for the reaction of calcium carbonate with dilute hydrochloric acid:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ ↑
(ii) Relative molecular mass of calcium carbonate=100
Mass of 4.5 moles of calcium carbonate
= No. of moles× Relative molecular mass
= 4.5×100
= 450g

76. (i) Calculate the volume of 320 g of SO₂ at STP (Atomic mass : S = 32 and O = 16)
(ii) State Gay-Lussac's Law of combining volumes.
(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 of propane (C₃H₅). (Atomic mass : C = 14, O = 16, H = 1, Molar Volume = 22.4 dm³ at STP)

Answer 

(i) Atomic mass: S = 32 and O = 16
Molecular mass of SO₂ = 32 + (2×16) = 64g
As 64 g of SO₂  = 22.4 dm³,

(ii) Gay-Lussac's law: When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure.
(iii) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
 Molar mass of propane = 44
44 g of propane requires 5 × 22.4 litres of oxygen at STP.

77. An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13% and Br = 85.11%. Find the molecule formula.
[Atomic mass C = 12, H = 1, Br = 80]

Answer 

Element	Relative atomic mass	% Compound	Atomic ratio	Simplest ratio
H	1	2.13	2.13/1 = 2.13	2
C	12	12.67	12.67/12 = 1.055	2
Br	80	85.11	85.11/80 = 1	1

Empirical formula = CH₂Br
n(Empirical formula mass of CH₂Br) = Molecular mass (2 × VD)
n(12 + 2 + 80) = 94 × 2
n = 2
Molecular formula = Empirical formula × 2
= (CH₂Br) × 2
= C₂H₄Br₂

78. Calculate the mass of
(i) 10²² atoms of sulphur
(ii) 0.1 mole of carbon dioxide
[Atomic mass : S = 32,C=12, and O = 16 and Avogadro's number = 6 × 10²³]

Answer 

(i) 10²² atoms of sulphur
6.022 × 10²³ atoms of sulphur will have mass = 32 g

(ii) 0.1 mole of carbon dioxide
1 mole of carbon dioxide will have mass = 44 g
0.1 mole of carbon dioxide will have mass = 4.4 g

79. O₂is evolved by heating KCIO₃ using MnO₂ as a catalyst.

(i) Calculate the mass of KClO₃ required to produce 6.72 litre of O₂at STP
[Atomic masses : K = 39, Cl = 35.5, O = 16]
(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules
(iii) Calculate the volume occupied by 0.01 mole of CO₂at STP

Answer 

80. Solve the following :
(i) What volume of oxygen is required to burn completely 90dm³ of butane under similar conditions of temperature and pressure? 
2C₄H₁₀ + 13O₂ → 8C₂ + 10H₂O
(ii) The vapour density of a gas is 8. What would be the volume occupied by 24.0 of the gas at STP?
(iii) A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?

Answer 

(iii) According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.
So, molecules of nitrogen gas present in the same vessel = x 

81. (i) Oxygen oxidase ethyne to carbon dioxide and water as shown by the equation :
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
What volume of ethylene gas at STP is required to produce 8.4 dm³of carbon dioxide at STP?
(ii) A compound made up of two elements X and Y has an empirical formula X₂Y. if the atomic weight of X is 10 and that Y is 5 and the compound has a vapour density 25, find its molecular formula.

Answer 

82. Give one word or phrase for the following:
(i) The ratio of the mass of a certain volume of gas to the mass of an equal volume of hydrogen under the same conditions of temperature and pressure.
(ii) Formation of ions from molecules

Answer 

(i) Vapour density
(ii) Ionisation 

83. (i) State Avogadro's Law
(ii) A cylinder contains 68g of ammonia gas at STP
(1) What is the volume occupied by this gas?
(2) How many moles of ammonia are present in the cylinder?
(3) How many molecules of ammonia are present in the cylinder? [N-14,H-1]

Answer 

(i) Avogadro's law: Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

(ii) (1) A cylinder contains 68 g of ammonia gas at STP √(a² + b² )
Molecular weight of ammonia = 17g/mole
68 g of ammonia gas at STP = ? 
1 mole = 22.4 dm³ 
∴ 4 mole = 22.4 × 4 = 89.6 dm³ 
(2) 4 moles of ammonia gas is present in the cylinder 
(3) 1 mole = 6.023 × 10²³ molecules 
4 moles = 24.092 × 10²³  molecules 

84. Which of the following would weigh the least?
(A) 2 gram atoms of Nitrogen
(B) 1 mole of Silver
(C) 22.4 litres of oxygen gas at 1 atmospheric pressure and 273K
(D) 6.02 × 10²³ atoms of carbon
[Atomic masses : Ag = 108, N - 14, O = 16,C = 12]
Answer 
(D) 6.02 × 10²³ atoms of carbon

85. Complete the following calculations. Show working for complete credit:
(i) Calculate the mass of calcium that will contain the same number of atoms as are present in 3.2 gm of Sulphur. [Atomic masses : S = 32, Ca = 40]
(ii) If 6 litres of hydrogen and 4 litres of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.
(iii) If the empirical formula of a compound is CH and it has a vapour density of 13, find the molecular formula of the compound.

Answer 

(i) 3.2 g of S has number of atoms = 6.023 × 10²³×3.2 /32
 = 0.6023×10²³
 So, 0.6023 × 10²³ atoms of Ca has mass=40×0.6023×10²³/6.023×10²³ = 4g
(ii) 6 litres of hydrogen and 4 litres of chlorine when mixed result in the formation of 8 litres of HCl gas.
When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leaving behind only 2 litres of hydrogen gas.
Therefore, the volume of the residual gas will be 2 litres.

(iii) For acetylene, molecular mass = 2 × V.D = 2 × 12 = 26 g 
The empirical mass = 12(c) + 1(H) = 13 g 
n = (molecular formula mass)/(Empirical formula weight) = 26/13 = 2 
Molecular formula of acetylene = 2 × Empirical formula =  C₂H₂ 
Similarly, for benzene molecular mass = 2 × V.D = 2 × 39 = 78 
 n = 78/13 = 6 
So, the molecular formula C₆H₆ 

86. Consider the following reaction and based on the reaction answer the questions that follow :

Calculate :
(i) The quantity in moles of (NH₄)₂Cr₂O₇ if 63 gm of (NH₄)₂Cr₂O₇ is heated.
(ii) The quantity in moles of nitrogen formed.
(iii) The volume in litres or dm³ of N₂evolved at STP
(iv) The mass in grams of Cr₂O₃ formed at the same time
[Atomic masses : H = 1, Cr=52, N=14]

Answer 

87. A gas cylinder contain 12 × 10²⁴ molecules of oxygen gas
If Avogadro's number is 6 × 10²³, calculate :
(i) The mass of oxygen present in the cylinder
(ii) The volume of oxygen at STP present in the Cylinder [O = 16]

Answer 

(i) 1 mole oxygen molecule = 6.022 × 10²³  molecules 
X moles of oxygen molecule = 12 × 10²⁴ molecules
19.93 moles of oxygen molecule = 12 × 10²⁴ molecules 

88. A gaseous hydrocarbon contains 82.76% of carbon. Given that its vapour density is 29, find its molecular formula.
[C = 12, H = 1]

Answer 

% of carbon = 82.76%
% of hydrogen = 100 - 82.76 = 17.24%

Element	% Weight	Atomic Weight	Relative No. of moles	Simplest Ratio
C	82.76	12	82.76/12 = 6.89	6.89/6.8 = 1 × 2 = 2
H	17.24	1	17.24/1 = 17.24	17.24/6.89 = 2.5 × 2 = 5

Empirical formula = C₂H₅
Empirical formula weight = 2 × 12 + 1 × 5 = 24 + 5 = 29
Vapour density = 29
Relative molecular mass = 29 × 2 = 58

89. The equation 4NH³5O₂ → 4NO + 6H₂O represents the catalytic oxidation of ammonia. If 100 cm³of ammonia is used, calculate the volume of oxygen required to oxidise the ammonia completely.

Answer 

90. A gas of 32 g mass has a volume of 20 litres at STP. Calculate the gram molecular weight of the gas.

Answer 

Mass of gas = 32 g
Volume occupied by 32g of gas = 20litres
No. of moles = Volume of gas/22.4 = 20/22.4 = 0.9 moles 
Gram molecular weight of gas = Mass of gas/No. of moles = 32/0.9 = 35.55 gm 
Gram molecular weight of gas = 35.55 g 

91. How much calcium oxide is formed when 82 g of calcium nitrate is heated? Also find the volume of nitrogen dioxide evolved:
2Ca(NO₃)₂→ 2CaO + 4NO₂ + O₂ (Ca = 40,N = 14, O = 16)

Answer 

2Ca(NO₃)₂→ 2CaO + 4NO₂ + O₂
Molecular weight of 2Ca(NO₃)₂ = 2[40+2(14+48) = 32g
Molecular weight of CaO = 2(40 + 16) = 112g
328 g of Ca(No₃]₂ liberates 4 moles of NO₂
328 g of Ca (NO₃)₂ liberates 4 × 22.4 L of NO₂

92. The ratio between the number of molecules in 2g of hydrogen and 32 g of oxygen is
(A) 1:2
(B) 1:0:01
(C) 1:1
(D) 0.01:1
[Given that H=1,O=16]

Answer 

Mole in 2 g of hydrogen = Given mass/Molar mass = 2/2 = 1 mol 
mole in 32 g of oxygen = Given mass/Molar mass = 32/16× 2 = 1 mol 
Hence the ratio = 1mol/1mol = 1 : 1 

93. (a) Calculate the number of gram atoms in 4.6 grams of sodium (Na = 23)
(b) Calculate the percentage of water of crystallization in CuSO_4. 5H₂O
(H = 1, O = 16, S = 32, Cu = 64)
(c) A compound of X and Y has the empirical formula XY₂. Its vapour density is equal to its empirical formula weight. Determine its molecular formula.

Answer 

Given : 
Wt. of the sodium = 4.6 g 
Mass of sodium atom = 23 
Number of gram moles of Na = Mass of sodium/Mass of sodium atom = 4.6/23 = 0.2 g 

94. (a) Propane burns in air according to the following equation :
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
What volume of propane is consumed on using 1000 cm³of air, considering only 20% of air contains oxygen?
(b) The mass of 11.2 litres of a certain gas at s.t.p is 24 g. Find gram molecular mass of the gas

Answer 

(a) Given:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Volume of air = 1000 cm³
Percentage of oxygen in air = 20%
From the given information,
According to Gay-Lussac's law,
1 vol. of propane consumes 5 vol. of oxygen.
Volume of oxygen = 1000 cm³ × 20% = 200 cm^3|
Therefore,
Volume of propane burnt for every 200 cm³ of oxygen,
40 cm³of propane is burnt.
(b) Given:
 Volume of gas at STP = 11.2 litres
Mass of gas at STP = 24 g
Gram molecular mass = ?
Mass of 22.4 L of a gas at STP is equal to its gram molecular mass.
11.2 L of the gas at STP weighs 24 g
So,
22.4 L of the gas will weigh
Gram molecular mass = 48 g 

95. A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.
(a) Find the number of moles of hydrogen present
(b) What weight of CO₂can the cylinder hold under similar conditions of temperature and pressure? (H = 1 C = 12, O = 16)
(c) If the number of molecules of hydrogen in the cylinder is X, calculate the number of CO₂ molecules in the cylinder with the same conditions of temperature and pressure.
(d) State the law that helped you to arrive at the above result

Answer 

(c) According to Avogadro's law, equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

(d) So, number of molecules of carbon dioxide in the cylinder = number of molecules of hydrogen in the cylinder = X