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

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

(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.

(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?

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?

(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.
(e) Relative molecular mass.

(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 NA.
(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?

(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

(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

(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.

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

(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.

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?

"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 C12.
(e) Avogadro's number is equal to ______.

(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.

(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.

(a) Na2SO4.10H2O.
(b) C6H12O6.

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

Given reaction is:
N + 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 ]

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 SO contains = 6 × 1023
= 10.2 × 1023 molecules.

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

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]

Pb + (N) + (O)
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)

(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]

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.

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)

(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.

(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

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].

Ferrous sulphate is FeSO4 . 7H2
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.

 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.

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.

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

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.

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.

(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 ?

(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]

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.

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'.

(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]

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.

 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)

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?

(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?

(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?

(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)

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?

(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
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)

(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]

 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?

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

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

(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.

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.

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)

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]

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?

(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

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)

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]

 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 ?

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.

(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]

(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?

(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]

(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

(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)

(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.

(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 ?

(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.

(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:
(b) Gay-Lussac's law
(c) Boyle's law
(d) Charle'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 ?

(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.

(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

(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.

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
(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.

(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.

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.

(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].

(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

(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)

(i) Atomic mass: S = 32 and O = 16
Molecular mass of SO= 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]

 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]

(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

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?

(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 :
2C2H+ 5O→ 4CO+ 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.

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

(i) Vapour density
(ii) Ionisation

(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]

(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]
(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.

(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]

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]

(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]

% 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 4NH35O→ 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.

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

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)

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]

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 CuSO4. 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.

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 :
C3H+ 5O→ 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

(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 cm3|
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