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

### Exercise 5 A

**Solution 1.**

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

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

**Solution 2.**

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

**means 1 molecule of nitrogen and 2N means two atoms of nitrogen.**

_{2 }N

**can exist independently but 2N cannot exist independently.**

_{2}**Solution 3.**

(a) This is due to Avogadros Law which states Equal volumes of all gases under similar conditions of temperature and pressure contain the same number of molecules.

Now the volume of hydrogen gas = volume of helium gas

n molecules of hydrogen = n molecules of helium gas

nH

**= nHe**

_{2}1 mol. of hydrogen has 2 atoms of hydrogen and 1 molecule of helium has 1 atom of helium

Therefore 2H=He

Therefore atoms in hydrogen is double the atoms of helium.

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

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

**Solution 4.**

From the equation, 2V of hydrogen reacts with 1V of oxygen

so 200cm

**of Hydrogen reacts with = 200/2= 100 cm**

^{3}

^{3}Hence, the unreacted oxygen is 150 – 100 = 50cm

**of oxygen.**

^{3}

**Solution 5:**

**reacts with = 2V of O**

_{4 }

_{2}so, 80 cm

**CH**

^{3}**reacts with = 80×2 = 160cm**

_{4}**O**

^{3}

_{2}Remaining O

**is 200-160 = 40cm**

_{2}

^{3}From the equation 1V of methane gives 1V of CO

_{2}So, 80 cm

**gives 80cm**

^{3}**CO**

^{3}**and H**

_{2}**O is negligible.**

_{2}**Solution 6:**

From equation, 2V of C

**H**

_{2}**requires = 5 V of O**

_{2 }

_{2}So, for 400ml C

**H**

_{2}**, O**

_{2}**required = 400 × 5/2 = 1000 ml**

_{2}Similarly, 2 V of C

**H**

_{2}**gives = 4 V of CO**

_{2}

_{2}So, 400ml of C

**H**

_{2}**gives CO**

_{2 }**= 400 × 4/2 = 800ml**

_{2}**Solution 7:**

Balanced chemical equation:

(i) At STP, 1 mole gas occupies 22.4 L.

As 1 mole H

**S gas produces 2 moles HCl gas,**

_{2}22.4 L H

**S gas produces 22.4 × 2 = 44.8 L HCl gas.**

_{2}Hence, 112 cm

**H**

^{3 }**S gas will produce 112 × 2 = 224 cm**

_{2}**HCl gas.**

^{3}**S gas consumes 1 mole Cl**

_{2}**gas.**

_{2}This means 22.4 L H

**S gas consumes 22.4 L Cl**

_{2}**gas at STP.**

_{2}Hence, 112 cm

**H**

^{3}**S gas consumes 112 cm**

_{2}**Cl**

^{3}**gas.**

_{2}120 cm

**- 112 cm**

^{3}**= 8 cm**

^{3}**Cl**

^{3}**gas remains unreacted.**

_{2}Thus, the composition of the resulting mixture is 224 cm

**HCl gas + 8 cm**

^{3}

^{3}Cl

**gas.**

_{2}**Solution 8:**

Now from equation, 2V of ethane reacts with = 7 V of oxygen

So, 600cc of ethane reacts with = 600 × 7/2 = 2100cc

Hence, unused O

**is = 2500 - 2100 = 400 cc**

_{2 }From 2V of ethane = 4 V of CO

**is produced**

_{2}So, 600cc of ethane will produce = (4 × 600/2) = 1200cc CO

_{2}**Solution 9:**

P

**V**

_{1}**/T**

_{1}**= P**

_{1}**V**

_{2}**/T**

_{2}

_{2}V

**= P**

_{2 }**V**

_{1}**T**

_{1}**/P**

_{2}**T**

_{2}

_{1}= (380 × 33 × 273)/(549 ×760) = 8.25 litres

**Solution 10:**

From equation, 1V of CH

**gives = 2V HCI**

_{4}so, 40 ml of methane gives 80 ml HCI

For 1V of methane = 2V of Cl

**required**

_{2}So, for 40ml of methane = 40 × 2 = 80 ml of Cl

_{2}**Solution 11.**

From equation, 5V of O

**required = 1V of propane**

_{2}So, 100 cm

**of O**

^{3}**will require = 20 cm**

_{2}**of propane**

^{3}**Solution 12:**

From equation, 1V of O

**reacts with = 2 V of NO**

_{2}200cm

**oxygen will react with = 200 × 2 = 400 cm**

^{3}**NO**

^{3}

^{3}NO

**produced = 400cm**

_{2}**because 1V oxygen gives 2 V NO**

^{3}

_{2}Total mixture = 400 + 50 = 450 cm

^{3}**Solution 13.**

2 V of CO requires = 1V of O

_{2}so, 100 litres of CO requires = 50 litres of O

_{2}**Solution 14:**

9 litres of reactants gives 4 litres of NO

So, 27 litres of reactants will give = 27 × 4/9 = 12 litres of NO

**Solution 15.**

Since 1 V hydrogen requires 1 V of oxygen and 4cm

**of H**

^{3}**remained behind so the mixture had com ">16 cm**

_{2}**hydrogen and 16 cm**

^{3}**chlorine.**

^{3}Therefore Resulting mixture is H

**= 4cm**

_{2}**, HCl = 32cm**

^{3}

^{3}**Solution 16.**

**From the equations, we can see that**

1V CH

**requires oxygen = 2V O**

_{4}

_{2}So, 10cm

**CH**

^{3}**will require = 20 cm**

_{4}**O**

^{3}

_{2}Similarly 2V C

**H**

_{2}**requires = 5V O**

_{2}

_{2}So, 10 cm

**C**

^{3}**H**

_{2}**will require = 25 cm**

_{2}**O**

^{3}

_{2}Now, 20 V O

**will be present in 100 V air and 25V O**

_{2}**will be present in 125 V air, so the volume of air required is 225 cm**

_{2}

^{3}**Solution 17:**

C

**H**

_{3}**+ 5O**

_{8}**+ 3CO**

_{2}**+ 4H**

_{2}**O**

_{2}2C

**H**

_{4}**+ 13O**

_{10}**→ 8CO**

_{2}**+ 10H**

_{2}**O**

_{2}60 ml of propane (C

**H**

_{3}**) gives 3 × 60 = 180 ml CO**

_{8}

_{2}40 ml of butane (C

**H**

_{4}**) gives = 8 × 40/2 = 160 ml of CO**

_{10}

_{2}Total carbon dioxide produced = 340 ml

So, when 10 litres of the mixture is burnt = 34 litres of CO

**is produced.**

_{2}**Solution 18:**

2C

**H**

_{2}**(g) + 5O**

_{2}**(g) → 4CO**

_{2}**(g) + 2H**

_{2}**O(g)**

_{2}4V CO

**is collected with 2 V C**

_{2}**H**

_{2}

_{2}So, 200m

**CO**

^{3}**will be collected with = 100cm**

_{2}**C**

^{3}**H**

_{2}

_{2}Similarly, 4V of CO

**is produced by 5 V of O**

_{2}

_{2}So, 200cm

**CO**

^{3}**will be produced by = 250 ml of O**

_{2}

_{2}**Solution 19:**

This experiment supports Gay lussac’s law of combining volumes.

Since the unchanged or remaining O

**is 58 cc so, used oxygen 106 - 58 = 48cc**

_{2}According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.

i.e. methane and oxygen react in a 1:2 ratio.

**Solution 20:**

**Solution 21:**

100 cm

**of oxygen contains = Y molecules**

^{3}Applying Avogadro's law,

50 cm

**nitrogen contains = 50Y/100 = Y/2**

^{3}**Exercise 5(B)**

**Solution 1:**

**a) This statement means one atom of chlorine is 35.5 times heavier than 1/12 time of the mass of an atom C-12.**

b) The value of Avogadro's number is 6.023 × 10

^{23}c) The molar volume of a gas at STP is 22.4 dm

**at STP**

^{3}**Solution 2:**

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

**.**

^{3}**atoms.**

^{23}(g) Mole is the amount of a substance containing elementary particles like atoms, molecules or ions in 12 g of carbon-12.

**Solution 3:**

(a) Applications of Avogadro’s Law :

1. It explains Gay-Lussac’s law.

2. It determines the atomicity of the gases.

3. It determines the molecular formula of a gas.

4. It determines the relation between molecular mass and vapour density.

5. It gives the relationship between gram molecular mass and gram molecular volume.

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

n molecules : n molecules : 2n molecules (By Avogadro's law)

**Solution 4:**

**(a) (2N)28 + (8H)8 + (Pt)195 + (6Cl)35.5×6 = 444**

(b) KClO

**= (K)39 + (Cl)35.5 + (3O)48 = 122.5**

_{3}(c) (Cu)63.5 + (S)32 + (4O)64 + (5H

**O)5×18 = 249.5**

_{2}(d) (2N)28 + (8H)8 + (S)32 + (4O)64 = 132

(e) (C)12 + (3H)3 + (C)12 + (2O)32 + (Na)23 = 82

(f) (C)12 + (H)1+ (3Cl)3×35.5 = 119.5

(g) (2N)28 + (8H)8 + (2Cr)2×51.9 + (7O)7×16 = 252

**Solution 5:**

**(a) No. of molecules in 73 g HCl = (6.023×10**

**×73)/36.5(mol. mass of HCl)**

^{23}= 12.04×10

^{23}**is = 32(mol. Mass of O**

_{2}**)×0.5=16 g**

_{2}**O = (6.023×10**

_{2}**×1.8)/18**

^{23 }= 6.023×10

^{22}**= 10/100(mol. Mass CaCO**

_{3}_{3})

= 0.1 mole

**gas = 2(Mol. Mass)×0.2 = 0.4 g**

_{2}**= (6.023×10**

_{2}**×3.2)/64**

^{23}= 3.023×10

^{22}**Solution 6:**

The molecular mass of H

**O is 18, CO**

_{2}**is 44, NH**

_{2}**is 17 and CO is 28**

_{3}So, the weight of 1 mole of CO

**is more than the other three.**

_{2}**Solution 7:**

**4g of NH**

**having minimum molecular mass contain maximum molecules.**

_{3}**Solution 8:**

a) No. of particles in s1 mole = 6.023 × 10

^{23}So, particles in 0.1 mole = 6.023 × 10

**× 0.1 = 6.023 × 10**

^{23}

^{22}b) 1 mole of H

**SO**

_{2}**contains = 2 × 6.023 × 10**

_{4}

^{23}So, 0.1 mole of H

**SO**

_{2}**contains = 2 × 6.023 × 10**

_{4}**× 0.1**

^{23}= 1.2 × 10

**atoms of hydrogen**

^{23}c) 111g CaCl

**contains = 6.023 × 10**

_{2}**molecules**

^{23}So, 1000 g contains = 5.42 × 10

**molecules**

^{24}**Solution 9:**

(a) 1 mole of aluminium has mass = 27 g

So, 0.2 mole of aluminium has mass = 0.2 × 27 = 5.4 g

(b) 0.1 mole of HCl has mass = 0.1 × 36.5 (mass of 1 mole)

= 3.65 g

(c) 0.2 mole of H

**O has mass = 0.2×18 = 3.6 g**

_{2}(d) 0.1 mole of CO

**has mass = 0.1×44 = 4.4 g**

_{2}**Solution 10:**

(a) 5.6 litres of gas at STP has mass = 12 g

So, 22.4 litre (molar volume) has mass = 12 × 22.4/5.6

= 48g(molar mass)

(b) 1 mole of SO

**has volume = 22.4 litres**

_{2}So, 2 moles will have = 22.4 × 2 = 44.8 litre

**Solution 11:**

(a) 1 mole of CO

**contains O**

_{2}**= 32g**

_{2}So, CO

**having 8 gm of O**

_{2}**has no. of moles = 8/32 = 0.25 moles**

_{2}(b) 16 g of methane has no. of moles = 1

So, 0.80 g of methane has no. of moles = 0.8/16 = 0.05 moles

**Solution 12:**

(a) 6.023×10

**atoms of oxygen has mass = 16 g**

^{23}So, 1 atom has mass = 16/6.023 × 10

**= 2.656 × 10**

^{23}**g**

^{-23}(b) 1 atom of Hydrogen has mass = 1/6.023 × 10

**= 1.666 × 10**

^{23}

^{-24}(c) 1 molecule of NH

**has mass = 17/6.023 × 10**

_{3}**= 2.82 × 10**

^{23}**g**

^{-23}(d) 1 atom of silver has mass = 108/6.023 ×10

**= 1.701 × 10**

^{23}

^{-22}(e) 1 molecule of O

**has mass = 32/6.023 × 10**

_{2}**= 5.314 × 10**

^{23}**g**

^{-23}(f) 0.25 gram atom of calcium has mass = 0.25 × 40 = 10g

**Solution 13:**

**(a) 0.1 mole of CaCO**

**has mass = 100(molar mass) × 0.1 = 10 g**

_{3}(b) 0.1 mole of Na

**SO**

_{2}**.10H**

_{4}**O has mass = 322 × 0.1 = 32.2 g**

_{2}(c) 0.1 mole of CaCl

**has mass = 111 × 0.1 = 11.1g**

_{2}(d) 0.1 mole of Mg has mass = 24 × 0.1 = 2.4 g

**Solution 14:**

**1molecule of Na**

**CO**

_{2}**.10H**

_{3}**O contains oxygen atoms = 13**

_{2}So, 6.023 ×10

**molecules (1mole) has atoms = 13 × 6.023 × 10**

^{23}

^{23}So, 0.1 mole will have atoms = 0.1 × 13 × 6.023 × 10

**= 7.8 × 10**

^{23}

^{23}**Solution 15:**

3.2 g of S has number of atoms = (6.023 × 10

**× 3.2)/32**

^{23}= 0.6023 × 10

^{23}So, 0.6023 × 10

**atoms of Ca has mass = (40 × 0.6023 × 10**

^{23}**)/(6.023 × 10**

^{23}**)**

^{23}

^{ }= 4g

**Solution 16:**

(a) No. of atoms = 52 × 6.023 × 10

**= 3.131 × 10**

^{23}

^{25}so, 52 amu = 13 atoms of He

(c) 4 g of He has atoms = 6.023 × 10

^{23}So, 52 g will have = (6.023 ×10

**× 52)/4 = 7.828 ×10**

^{23}**atoms**

^{24}**Solution 17:**

Molecular mass of Na

**CO**

_{2}**= 106 g**

_{3 }106 g has 2 x 6.023 x 10

**atoms of Na**

^{23}So, 5.3g will have = (2 × 6.023 × 10

**× 5.3)/106 = (6.022 × 10**

^{23}**)atoms**

^{22}Number of atoms of C = (6.023 ×10

**×5.3)/106 = 3.01 ×10**

^{23}**atoms**

^{22}And atoms of O = (3 ×6.023 ×10

**×5.3)/106 = 9.03×10**

^{23}**atoms**

^{22}**Solution 18:**

(a) 60 g urea has mass of nitrogen(N

**) = 28 g**

_{2}So, 5000 g urea will have mass = 28 × 5000/60 = 2.33 kg

(b) 64 g has volume = 22.4 litre

So, 320 g will have volume = 22.4×320/64 = 112 litres

**Solution 19:**

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

(b) Vapour density of Chlorine atom is 35.5.

**Solution 20:**

22400 cm

**of CO has mass = 28 g**

^{3}So, 56 cm

**will have mass = 56 ×28/22400 = 0.07 g**

^{3}**Solution 21:**

18 g of water has number of molecules = 6.023 ×10

^{23}So, 0.09 g of water will have no. of molecules = (6.023× 10

**×0.09)/18 = 3.01 × 10**

^{23}**molecules**

^{21}**Solution 22:**

(a) No. of moles in 256 g S

**= 1 mole**

_{8}So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles

(b) No. of molecules = 0.02 × 6.023 × 10

**= 1.2 × 10**

^{23}**molecules**

^{22}No. of atoms in 1 molecule of S = 8

So, no. of atoms in 1.2 × 10

**molecules = 1.2 × 10**

^{22}**× 8**

^{22}= 9.635 × 10

**molecules**

^{22}**Solution 23:**

The atomic mass of phosphorus P = 30.97 g

Hence, the molar mass of P

**= 123.88 g**

_{4}If phosphorus is considered as P

**molecules,**

_{4}then 1 mole P

**≡ 123.88 g**

_{4}Therefore, 100 g of P

**= 0.807 g**

_{4}**Solution 24:**

(a) 308 cm

**of chlorine weighs = 0.979g**

^{3}So, 22400 cm

**will weigh = gram molecular mass**

^{3}= 0.979 ×22400/308 = 71.2 g

**at 1 atm has volume = 22.4 litres**

_{2}So, 4g H

**at 1 atm will have volume = 44.8 litres**

_{2}Now, at 1 atm(P

**) 4g H**

_{1}**has volume (V**

_{2}**) = 44.8 litres**

_{1}So, at 4 atm(P

**) the volume(V**

_{2}**) will be P**

_{2}**V**

_{1}**/P**

_{1}**= (1 × 44.8)/4 = 11.2 litres**

_{2}So, mass of oxygen in 2.2 litres = (2.2 × 32)/22.4 = 3.14 g

**Solution 25:**

**No. of atoms in 12 g C = 6.023 × 10**

^{23}So, no. of carbon atoms in 10

**g = (10**

^{-12}**× 6.023 × 10**

^{-12}**)/12**

^{23}= 5.019 × 10

**atoms**

^{10}**Solution 26:**

Given:

P = 1140 mm Hg

Density = D = 2.4 g / L

T = 273°C = 273 + 273 = 546 K

M = ?

We know that, at STP, the volume of one mole of any gas is 22.4 L

Hence we have to find out the volume of the unknown gas at STP.

First, apply Charle’s law.

We have to find out the volume of one litre of an unknown gas at a standard temperature of 273 K.

V

**= 1 L, T**

_{1}**= 546 K**

_{1}V

**= ?, T**

_{2}**= 273 K**

_{2 }V

**/T**

_{1}**= V**

_{1}**/ T**

_{2}

_{2}V

**= (V**

_{2}**× T**

_{1}**)/T**

_{2}

_{1}= 0.5 L

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

Apply Boyle’s law.

P

**= 1140 mm Hg, V**

_{1}**= 0.5 L**

_{1}P

**= 760 mm Hg, V**

_{2}**= ?**

_{2}P

**× V**

_{1}**= P**

_{1}**× V**

_{2}

_{2}V

**= (P**

_{2}**× V**

_{1}**)/P**

_{1}

_{2}= (1140 mm Hg × 0.5 L)/760 mm Hg

= 0.75 L

X moles = 0.75 L/22.4 L

= 0.0335 moles

The original mass is 2.4 g

n = m / M

0.0335 moles = 2.4 g / M

M = 2.4 g/0.0335 moles

M = 71.6 g/mole

Hence, the gram molecular mass of the unknown gas is 71.6 g

**Solution 27:**

1000 g of sugar costs = Rs. 40

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

**Solution 28:**

(a) Weight of 1 g atom N = 14 g

So, weight of 2 g atom of N = 28 g

**atoms of C weigh = 12 g**

^{23}So, 3 x 10

**atoms will weigh = (12 × 3 × 10**

^{25}**)/(6.023 × 10**

^{25}**) = 597.79**

^{23}So, 7 grams of silver weighs least.

**Solution 29:**

40 g of NaOH contains 6.023 × 10

**molecules**

^{23}So, 4 g of NaOH contains = (6.02 × 10

**× 4)/40**

^{23}= 6.02 ×10

**molecules**

^{22}**Solution 30:**

**The number of molecules in 18 g of ammonia = 6.02 ×10**

^{23}So, no. of molecules in 4.25 g of ammonia = (6.02 ×10

**×4.25)/18**

^{23}= 1.5 × 10

^{23}

**Solution 31:**

**(a) One mole of chlorine contains 6.023 ×10**

**atoms of chlorine.**

^{23}(b) Under similar conditions of temperature and pressure, two volumes of hydrogen combined with one volume of oxygen will give two volumes of water vapour.

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

(d) Under similar conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

**Exercise 5 C**

**Solution 1:**

**Information conveyed by H**

**O**

_{2}1. That H

**O contains 2 volumes of hydrogen and 1 volume of oxygen.**

_{2}2. That ratio by weight of hydrogen and oxygen is 1:8.

3. That molecular weight of H

**O is 18g.**

_{2}

**Solution 2:**

**The empirical formula is the simplest formula, which gives the simplest ratio in whole numbers of atoms of different elements present in one molecule of the compound.**

The molecular formula of a compound denotes the actual number of atoms of different elements present in one molecule of a compound.

**Solution 3.**

(a) CH

**O**

_{2}**O**

_{2}

**Solution 4:**

**Relative mol. mass of CuSO**

**.5H**

_{4}**O = 63.5 + 32 + (16 × 4) + 5(1×2 + 16)**

_{2}= 249.5 g

249.5 g of CuSO

**.5H**

_{4}**O contains water of crystallization = 90 g**

_{2}So, 100 g will contain = (90 ×100)/249.5 = 36.07 g

So, % of H

**O = 36.07 × 100 = 36.07%**

_{2}**Solution 5:**

**(a) Molecular mass of Ca(H**

**PO**

_{2}**)**

_{4}**= 234**

_{2}So, % of P = (2 × 31 × 100)/234 = 26.5%

**(PO**

_{3}**)**

_{4}**= 310**

_{2}% of P = (2 × 31 × 100)/310 = 20%

**Solution 6:**

**Molecular mass of KClO**

**= 122.5 g**

_{3}% of K = 39 /122.5 = 31.8%

% of Cl = 35.5/122.5 = 28.98%

% of O = 3 × 16/122.5 = 39.18%

**Solution 7:**

Element |
% |
At. mass |
Atomic ratio |
Simple ratio |

Pb |
62.5 |
207 |
62.5/207 |
0.3019 |

N |
8.5 |
14 |
8.5/14 |
0.6071 |

O |
29.0 |
16 |
29.0/16 |
1.81 |

**)**

_{3}**is the empirical formula.**

_{2}**Solution 8:**

In Fe

**O**

_{2}**, Fe = 56 and 0 = 16**

_{3}Molecular mass of Fe

**O**

_{2}**= 2 × 56 + 3 × 16 = 160 g**

_{3}Iron present in 80% of Fe

_{2}O

**= 112/(160 × 80) = 56g**

_{3}So, mass of iron in 100 g of ore = 56 g

∴ mass of Fe in 10000 g of ore = (56 × 10000)/100

**Solution 9:**

**For acetylene, molecular mass = 2 × V.D = 2 × 13 = 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 = C

**H**

_{2}

_{2}Similarly, for benzene molecular mass = 2 × V.D = 2 × 39 = 78

n = 78/13 = 6

So, the molecular formula = C

**H**

_{6}

_{6}

**Solution 10:**

Element |
% |
At. mass |
Atomic ratio |
Simple ratio |

H |
17.71 |
1 |
17.7/1 = 17.7 |
17.7/5.87 = 3 |

N |
82.3 |
14 |
82.3/14 = 5.87 |
5.87/5.87 = 1 |

_{3}**Solution 11:**

Element |
% |
At. mass |
Atomic ratio |
Simple ratio |

C |
54.54 |
12 |
54.54/12 |
4.55 |

H |
9.09 |
1 |
9.09/1 |
9.09 |

O |
36.36 |
16 |
36.36/16 |
2.27 |

**H**

_{2}**O**

_{4}(b) empirical formula mass = 44

Since, vapour density = 44

So, molecular mass = 2 × V.D = 88

Or, n = 2

So, molecular formula = (C

**H**

_{2}**O)**

_{4}**= C**

_{2}**H**

_{4}**O**

_{8}

_{2}**Solution 12:**

Element |
% |
At. mass |
Atomic ratio |
Simple ratio |

C |
26.59 |
12 |
26.59/12 |
2.21 |

H |
2.22 |
1 |
2.22/1 |
2.22 |

O |
71.19 |
16 |
71.19/16 |
4.44 |

_{2}(b) empirical formula mass = 45

Vapour density = 45

So, molecular mass = V.D × 2 = 90

So, molecular formula = C

**H**

_{2}**O**

_{2}

_{4}**Solution 13:**

Element |
% |
At. mass |
Atomic ratio |
Simple ratio |

Cl |
71.65 |
35.5 |
71.65/35.5 |
2.01 |

H |
4.07 |
1 |
4.07/1 |
4.07 |

C |
24.28 |
12 |
24.28/12 |
2.02 |

**Cl**

_{2}(b) empirical formula mass = 49.5

Since, molecular mass = 98.96

so, molecular formula = (CH

**Cl)**

_{2}**= C**

_{2}**H**

_{2}**Cl**

_{4}

_{2}

**Solution 14:**

(a) The g atom of carbon = 4.8/12 = 0.4 and g atom of hydrogen = 1/1 = 1

(b)

Element |
Given mass |
At. mass |
Gram atom |
Ratio |

C |
4.8 |
12 |
0.4 |
12 |

H |
1 |
1 |
1 |
5 |

**H**

_{2}

_{5}(c) Empirical formula mass = 29

Molecular mass = V.D × 2 = 29 × 2 = 58

So, molecular formula = C

**H**

_{4}

_{10}**Solution 15:**

Since, g atom of Si = given mass/mol. Mass

So, given mass = 0.2 × 28 = 5.6 g

Element |
mass |
At. mass |
Gram atom |
Ratio |

Si |
5.6 |
28 |
0.2 |
1 |

Cl |
21.3 |
35.5 |
21.3/35.5 |
0.6 |

_{3}**Solution 16:**

Element |
mass |
At. mass |
Gram atom |
Ratio |

C |
92.3 |
12 |
92.3/12 |
7.7 |

H |
7.7 |
1 |
7.7/1 |
7.7 |

Empirical formula mass = 13

Since molecular mass = 78

So, n = 6

∴ molecular formula is C

**H**

_{6}

_{6}**Solution 17:**

**(a) G atoms of magnesium = 18/24 = 0.75 or g- atom of Mg**

(b) G atoms of nitrogen = 7/14 = 0.5 or 1/2 g- atoms of N

(c) Ratio of gram-atoms of N and Mg = 1:1.5 or 2:3

So, the formula is Mg

**N**

_{3}

_{2}**Solution 18:**

Barium chloride = BaCl

**.x H**

_{2}**O**

_{2}Ba + 2Cl + x[H

**+ O]**

_{2}= 137+ 235.5 + x[2+16]

= [208 + 18x] contains water = 14.8% water in BaCl

**.xH**

_{2}**O**

_{2}= [208 + 18x] 14.8/100 = 18x

= [104 + 9x] 2148 = 18000x

= [104+9x] 37 = 250x

= 3848 + 333x = 2250x

1917x = 3848

x = 2 molecules of water

**Solution 19:**

**Molar mass of urea; CON**

**H**

_{2}**= 60 g**

_{4}So, % of Nitrogen = 28 × 100/60 = 46.66%

**Solution 20:**

Element |
% |
At. mass |
Atomic ratio |
Simple ratio |

C |
42.1 |
12 |
3.5 |
1 |

H |
6.48 |
1 |
6.48 |
2 |

O |
51.42 |
16 |
3.2 |
1 |

**O**

_{2}Since the compound has 12 atoms of carbon, so the formula is

C

**H**

_{12}**O**

_{24}

_{12.}

**Solution 21:**

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

**B**

_{2}**.**

_{4}(b) Since molecular mass is 2 times the vapour density, so Mol. Mass = 2 V.D

Empirical formula weight = V.D/3

So, n = molecular mass/ Empirical formula weight = 6

Hence, the molecular formula is A

**B**

_{6}

_{6}

**Solution 22:**

**Atomic ratio of N = 87.5/14 = 6.25**

Atomic ratio of H = 12.5/1 = 12.5

This gives us the simplest ratio as 1:2

So, the molecular formula is NH

_{2}**Solution 23:**

Element |
% |
At. mass |
Atomic ratio |
Simple ratio |

Zn |
22.65 |
65 |
0.348 |
1 |

H |
4.88 |
1 |
4.88 |
14 |

S |
11.15 |
32 |
0.348 |
1 |

O |
61.32 |
16 |
3.83 |
11 |

**O**

_{14}

_{11}Empirical formula mass = 65.37+ 32 + 141 + 11 + 16 = 287.37

Molecular mass = 287

n = Molecular mass/Empirical formula mass = 287/287=1

Molecular formula = ZnSO

**H**

_{11}

_{14}= ZnSO

**.7H**

_{4}**O**

_{2}**Exercise 5 D**

**Solution 1:**

**(a) Moles: 1 mole + 2 mole → 1 mole + 2 mole**

(b) Grams: 42g + 36g → 745 +4g

(c) Molecules = 6.02 × 10

**+ 12.046 × 10**

^{23}**→ 6.02 × 10**

^{23}**+ 12.046 × 10**

^{23}

^{23}**Solution 2:**

(a)100 g of CaCO

**produces = 164 g of Ca(NO**

_{3}**)**

_{3}

_{2}So, 15 g CaCO

**will produce = (164 × 15/100) = 24.6 g Ca(NO**

_{3}**)**

_{3}

_{2}(b) 1 V of CaCO

**produces 1 V of CO**

_{3}

_{2}100 g of CaCO

**has volume = 22.4 litres**

_{3}So, 15 g will have volume = (22.4 × 15/100) = 3.36 litres CO

_{2}**Solution 3:**

2NH

**+ H**

_{3}**SO**

_{2}**→ (NH**

_{4}**)**

_{4}**SO**

_{2}

_{4}66 g

(a) 2NH

**+ H**

_{3}**SO4 → (NH**

_{2}**)**

_{4}**SO**

_{2}

_{4}For 132 g (NH

**)**

_{4}**SO**

_{2}**= 34 g of NH**

_{4}**is required**

_{3}So, for 66 g (NH

**)**

_{4}**SO**

_{2}**= (66 × 32/132) = 17 g of NH**

_{4}**is required**

_{3}(b) 17g of NH

**requires volume = 22.4 litres**

_{3}(c) Mass of acid required, for producing 132g (NH

**)**

_{4}**SO**

_{2}**= 98g**

_{4}So, Mass of acid required, for 66g (NH

**)**

_{4}**SO**

_{2}**= (66 × 98/132) = 49g**

_{4}**Solution 4:**

(a) Molecular mass of Pb

**O**

_{3}**= (3 × 207.2 + 4×16) = 685 g**

_{4}685 g of Pb

**O**

_{3}**gives = 834 g of PbCl**

_{4}

_{2}Hence, 6.85 g of Pb

**O**

_{3}**will give = (6.85 × 834)/685 = 8.34 g**

_{4}**O**

_{3}**gives = 71g of Cl**

_{4}

_{2}Hence, 6.85 g of Pb

**O**

_{3}**will give = (6.85 × 71)/685 = 0.71 g Cl**

_{4}

_{2}**O**

_{3}**produces 1 V Cl**

_{4 }

_{2}685g of Pb

**O**

_{3}**has volume = 22.4 litres = volume of Cl**

_{4}**produced**

_{2}So, 6.85 Pb

**O**

_{3}**will produce = (6.85 × 22.4/685) = 0.224 litres of Cl**

_{4}

_{2}**Solution 5:**

**The molecular mass of KNO**

_{3}= 101 g

63 g of HNO

**is formed by = 101 g of KNO**

_{3}

_{3}So, 126000 g of HNO

**is formed by = (126000 × 101)/63 = 202 kg**

_{3}Similarly, 126 g of HNO

**is formed by 170 kg of NaNO**

_{3 }

_{3}So, a smaller mass of NaNO

**is required.**

_{3}**Solution 6:**

CaCO

**+ 2HCI → CaCl**

_{3}**+ H**

_{2}**O + CO**

_{2}

_{2}(a) V

**= 2 litres V**

_{1}**= ?**

_{2 }T

**= (273 + 27) = 300K T**

_{1}**= 273K**

_{2}V

**/T**

_{1}**= V**

_{1 }**/T**

_{2}

_{2}V

**= V**

_{2}**T**

_{1}**/T**

_{2}**= [2 × 273/300]L**

_{1 }Now at STP 22.4 litres of CO

**are produced using CaCO**

_{2}**= 100g**

_{3}So, [2 × 273/300] litres are produced by = 100/22.4 2274/300 = 125g

(b) 22.4 litres are CO

**are prepared from acid = 73g**

_{2}[2 × 273/300] litres are prepared from = (73/22.4) × (2×273/300) = 5.9g

**Solution 7:**

**2 moles of H**

**O gives = 1 mole of O**

_{2}

_{2}So, 1 mole of H

**O will give = 0.5 moles of O**

_{2}

_{2}So, mass of O

**= no. of moles x molecular mass**

_{2}= 0.5 × 32 = 16 g of O

_{2}and 1 mole of O

**occupies volume = 22.4 litre**

_{2}So, 0.5 moles will occupy = 22.4 × 0.5 = 11.2 litres at S.T.P.

**Solution 8:**

(a) Mol. Mass of Na

**O**

_{2}**= (2 × 23 + 2 × 16) = 78 g**

_{2}Mass of 2Na

**O**

_{2}**= 156 g**

_{2}156 g Na

**O**

_{2}**gives = 160 g of NaOH (4 × 40 g)**

_{2 }So, 1.56 Na

**O**

_{2}**will give = (160 × 1.56/156) = 1.6 g**

_{2}**O**

_{2}**gives = 22.4 litres of oxygen**

_{2}So, 1.56 g will give = (22.4 × 1.56/156) = 0.224 litres = 224 cm

^{3}**O**

_{2}**gives = 32 g O**

_{2}

_{2}So, 1.56 g Na

**O**

_{2}**will give = (32 × 1.56/156)**

_{2 }= 32/100 = 0.32 g

**Solution 9:**

**Cl = 2(14 + (1 × 4) + 35.5] = 2[53.5] = 107 g**

_{4}(a) 107 g NH

**Cl gives = 34 g NH**

_{4}

_{3}So, 21.4 g NH

**Cl will give = (21.4 × 34/107) = 6.8 g NH**

_{4}

_{3}(b) The volume of 17 g NH

**is 22.4 litre**

_{3}So, volume of 6.8 g will be = 6.8 × 22.4/17 = 8.96 litre

**Solution 10:**

Now, since 144 g of Al

**C**

_{4}**gives = 3 × 22.4 litre of CH**

_{3}

_{4}So, 14.4 g of Al

**C**

_{4}**willgive = (3 × 22.4 × 14.4)/144 = 6.72 litres CH**

_{3}

_{4}**Solution 11:**

**(a) 1 mole of MnO**

**weighs = 87 g (mol. Mass)**

_{2}So, 0.02 mole will weigh = 87 × 0.02 = 1.74 g MnO

_{2}**gives = 1 mole of MnCl**

_{2}

_{2}So, 0.02 mole MnO

**will give = 0.02 mole of MnCl**

_{2}

_{2}**weighs = 126 (g/mol mass)**

_{2}So, 0.02 mole MnCl

**will weigh = 126 × 0.02 g = 2.52 g**

_{2}**will form = 0.02 mole of Cl**

_{2}

_{2}**weighs = 35.5 g**

_{2 }So, 0.02 mole will weigh = 71 × 0.02 = 1.42 g of Cl

_{2}So, 0.02 mole will have volume = (22.4 × 0.02) = 0.448 litre

**requires HCl = 4 mole**

_{2}So, 0.02 mole MnO

**will require = 4 × 0.02 = 0.08 mole**

_{2}**, acid required = 4 mole of HCI**

_{2}So, for 0.02 mole, acid required = 4 × 0.02 =0.08 mole

Mass of HCI = 0.08 × 36.5 = 2.92 g

**Solution 12:**

28g of nitrogen requires hydrogen = 6g

2000g of nitrogen requires hydrogen = 6/28 × 2000 = 3000/7g

So mass of hydrogen left unreacted = 1000 - 3000/7 = 571.4g of H

_{2}**= 34g**

_{3}2000g of N

**forms NH**

_{2}

_{3}= 34/28 × 2000

= 2428.6g

**Miscellaneous Exercise**

**Solution 1:**

**From equation: 2H**

**+ O**

_{2}**→ 2H**

_{2}**O**

_{2}1 mole of Oxygen gives = 2 moles of steam

So, 0.5 mole oxygen will give = 2 × 0.5 = 1mole of steam

**Solution 2:**

Mol. Mass of 8HNO

**= 8 × 63 = 504 g**

_{3}(a) For 504 g HNO

**, Cu required is = 192 g**

_{3}So, for 63g HNO

**Cu required = (192 × 63/504) = 24g**

_{3}**gives = 2 × 22.4 litre volume of NO**

_{3}So, 63g of HNO

**gives = (2 x 22.4 × 63)/504 = 5.6 litre of NO**

_{3}**Solution 3:**

**(a) 28g of nitrogen = 1mole**

So, 7g of nitrogen = 1/28 × 7 = 0.25 moles

**at STP = 22.4 litres**

_{2}Volume of 7.1 g chlorine = (22.4 × 7.1)/71 = 2.24 litre

(c) 22400cm

**volume have mass = 28 g of CO(molar mass)**

^{3}So, 56cm

**volume will have mass = (28 × 56)/22400 = 0.07 g**

^{3}**Solution 4:**

% of Nin NaNO = (14/85 × 100) = 16.47%

% of Nin (NH

**)**

_{4}**SO**

_{2}**= (14/132 × 100) = 21.21%**

_{4 }% of N in CO(NH

**)**

_{2}**= (14/60 × 100) = 46.66%**

_{2}So, highest percentage of N is in urea.

**Solution 5:**

(a) From equation, 2V of water gives 2V of H

**and 1 V of O**

_{2}

_{2}where 2 V = 2500 cm

^{3}so, volume of O

**liberated = 2V/V = 1250 cm**

_{2}

^{3}**V**

_{1}**/T**

_{1}**= P**

_{1}**V**

_{2}**/T**

_{2}

_{2}P

**V**

_{1}**/T**

_{1}**= (7P**

_{1}**× V**

_{1}**)/(2 × T**

_{2}**)**

_{1}V

**= (2500×2)/7**

_{2}V

**= 5000/7 cm**

_{2}

^{3}**/V**

_{1}**= T**

_{2}**/T**

_{1}

_{2}5000/(7 × 2500) = (T

**/T**

_{1}**)**

_{2}T

**= 3.5 T**

_{2}

_{1}i.e. temperature should be increased by 3.5 times.

**Solution 6:**

**Molecular mass of urea = 12 + 16 + 2(14 + 2) = 60g**

60g of urea contains nitrogen = 28g

So, in 50g of urea, nitrogen present = 23.33 g

50 kg of urea contains nitrogen = 23.33kg

**Solution 7:**

(a) 80% C and 20% H

So, atomic ratio of C and H are: C = 80/12 = 6.66; H = 20/1 = 20

Simple ratio of C:H = 1: 3

So, empirical formula is CH

_{3}Vapour density = 15

So, the molecular mass = 15(V.D) × 2 = 30 g

Hence, n = 2 so the molecular formula is C

**H**

_{2}

_{3}**Solution 8:**

22400 cm

**CO**

^{3}**has mass = 44g**

_{2}So, 224 cm

**CO**

^{3}**will have mass = 0.44g**

_{2}Now since CO

**is being formed and X is a hydrocarbon so it contains C and H.**

_{2}In 0.44g CO

**, mass of carbon = 0.44 - 0.32 = 0.12g = 0.012 g atom**

_{2}So, mass of Hydrogen in X = 0.145 - 0.12 = 0.025g = 0.025g atom

Now the ratio of C : H is C = 1:H = 2.5 or C = 2 : H = 5

i.e. the formula of hydrocarbon is C

**H**

_{2}

_{5}(a) C and H

(b) Copper (11) oxide was used for reduction of the hydrocarbon.

(c) (i) no. of moles of CO

**= 0.44/44 = 0.01 moles**

_{2}(ii) mass of C = 0.12g

(iii) mass of H = 0.025 g

(iv) The empirical formula of X = C

**H**

_{2}

_{5}**Solution 9:**

**Mass of X in the given compound = 24g**

Mass of oxygen in the given compound = 64g

So total mass of the compound = 24 + 64 = 88g

% of X in the compound = 24/88 100 = 27.3%

% of oxygen in the compound = 64/88 100 = 72.7%

Element |
% |
At. Mass |
Atomic ratio |
Simplest ratio |

X |
27.3 |
12 |
27.3/12 |
2.27 |

O |
72.7 |
16 |
72.2/16 |
4.54 |

_{2}**Solution 10:**

(a) V.D = mass of gas at STP/mass of equal volume of H

**= 85/5 = 17**

_{2}(b) Molecular mass = 17(V.D) × 2 = 34g

**Solution 11:**

(a) 12 g of C gives = 44.8 litre volume of CO

So, 3 g of C will give = 11.2 litre of CO

so, 24 cm

**CO will require = 24/2 = 12 cm**

^{3}

^{3}(ii) 2 x 22400 cm

**CO gives = 2 × 22400 cm**

^{3}**CO**

^{3}

_{2}So, 24cm

**CO will give = 24 cm**

^{3}**CO**

^{3}

_{2}**Solution 12:**

(a) 56 g of CaO is obtained with NO

**= 2 × 22.4 litre of NO**

_{2}

_{2}So, 5.6g of CaO is obtained with NO

**= (2 × 22.4 × 5.6/56) = 4.48 litre**

_{2}**)**

_{3}

_{2}So, 5.6 g Cao is obtained by = 5.6 × 56/164 g Ca(NO

**)**

_{3}

_{2}= 16.4 g of Ca(NO

**)**

_{3}**is heated.**

_{2}**Solution 13:**

(a) Number of molecules in 100cm

**of oxygen = Y**

^{3}According to Avogadro's law, Equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of molecules. Therefore, the number of molecules in 100 cm

**of nitrogen under the same conditions of temperature and pressure = Y**

^{3}So, the number of molecules in 50 cm

**of nitrogen under the same conditions of temperature and pressure = Y/100 50 = Y/2**

^{3}(ii) The empirical formula is CH

_{3}(iii) The empirical formula mass for CH

**O = 30**

_{2}V.D = 30

Molecular formula mass = V.D 2 = 60

Hence, n = mol. Formula mass/empirical formula mass = 2

So, molecular formula = (CH

**O)**

_{2}**= C**

_{2}**H**

_{2}**O**

_{4}

_{2}**Solution 14:**

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

**Solution 15:**

Mass of silicon in the given compound = 5.6g

Mass of the chlorine in the given compound = 21.3g

Total mass of the compound = 5.6g + 21.3g = 26.9g

% of silicon in the compound = (56/26.9 ×100) = 20.82%

% of chlorine in the compound = (21.2/26.9 ×100) = 79.18%

Element |
% |
At. Mass |
At. Ratio |
Simplest ratio |

Si |
20.82 |
28 |
20.82/28 |
0.741 |

Cl |
79.18 |
35.5 |
79.18/35.5 |
2.233 |

_{3}**Solution 16:**

Element |
% composition |
Atomic ratio |
Simple ratio |

P |
38.27 |
38.27/31 |
1.23 |

H |
2.47 |
2.47/1 |
2.47 |

O |
59.26 |
59.26/16 |
3.703 |

So, empirical formula is PH

**O**

_{2}**or H**

_{3}**PO**

_{2}

_{3}Empirical formula mass = (31 + 2×1 + 3×16) = 81

The molecular formula is = H

**P**

_{4}**O**

_{2}**, because n = 162/81 = 2**

_{6}

**Solution 17:**

V

**= 10 litres V**

_{1}**= ?**

_{2}T

**= 27+ 273 = 300K T**

_{1}**= 273K**

_{2}P

**=700 mm P**

_{1}**= 760 mm**

_{2}Using the gas equation,

P

**V**

_{1}**/T**

_{1}**= P**

_{1}**V**

_{2}**/T**

_{2}

_{2}V

**= P**

_{2}**V**

_{1}**T**

_{1}**/T**

_{2}**P**

_{1}**= (700 × 10 × 273)/(300 × 760)**

_{2}Molecular weight A = 60

So, weight of 22.4 litres of A at STP = 60 g

Weight of = (700 × 10 × 273)/(300 × 760) litres of A at STP

= 60/22.4 × (700 × 10 × 273/300 × 760) g or 22.45g

**Solution 18:**

(a) Molecular mass of CO

**= 12 + 2×16 = 44 g**

_{2}So, vapour density (V.D) = mol. Mass/2 = 44/2 = 22

V.D = mass of certain amount of CO

**/mass of equal volume of hydrogen = m/1**

_{2}22 = m/1

So, mass of CO

**= 22 kg**

_{2}So, number of molecules of carbon dioxide in the cylinder = number of molecules of hydrogen in the cylinder = X

**Solution 19:**

**(a) The volume occupied by 1 mole of chlorine = 22.4 litre**

**/V**

_{1}**= T**

_{2}**/T**

_{1}

_{2}22.4/V

**= 273/546**

_{2}V

**= 44.8 litres**

_{2}**gas = 35.5 × 2 = 71 g**

_{2}

**Solution 20:**

**(a) Total molar mass of hydrated CaSO**

**.xH**

_{4}**O = 136 + 18x**

_{2}Since 21% is water of crystallization, so

18x/(136 + 18x) = 21/100

So, x = 2 i.e. water of crystallization is 2.

So, for 1.8 g, vol. of H

**needed = (1.8 × 22.4/18) = 2.24 litre**

_{2}Now 2 vols. of water = 1 vol. of oxygen

1 vol. of water = 1/2 vol. of O

**= 22.4/2 = 11.2 lit.**

_{2}18 g of water = 11.2 lit. of O

_{2}1.8 g of water = 11.2/18 18/10 = 1.12 lit.

2g will occupy = 224002/32 = 1400cc

P

**= 760mm P**

_{1}**= 740mm**

_{2}V

**= 1400cc V**

_{1}**= ?**

_{2 }T

**= 273K, T**

_{1}**= 27 + 73 = 300K**

_{2}P

**V**

_{1}**/T**

_{1}

_{1}**= P**

_{ }**V**

_{2}**/T**

_{2}

_{2}V

**= P**

_{2}**V**

_{1}**T**

_{1}**/T**

_{2}**V**

_{1}**= (760 × 1400 × 300)/(273 × 740) = 1580cc**

_{2}1580/1000 = 1.581

**= 750mm P**

_{1}**= 760mm**

_{2}V

**= 44lit. V**

_{1}**= ?**

_{2}T

**= 298K T**

_{1}**= 273K**

_{2}P

**V**

_{1}**/T**

_{1}**= P**

_{1}**V**

_{2}**/T**

_{2}

_{2}V

**= P**

_{2}**V**

_{1}**T**

_{1}**/T**

_{2}**P**

_{1}**= (750 × 44 × 273)/(298 × 760) = 39.78 lit.**

_{2}22.4 lit. of CO

**at STP has mass = 44g**

_{2}39.78 lit. of CO

**at STP has masss = (44 × 39.78)/22.4 = 78.14 g**

_{2}so, 1.435 g of AgCl is formed by = 0.585 g of NaCl

% of NaCl = 0.585 × 100 = 58.5%

**Solution 21:**

P

**V**

_{1}**/T**

_{1}**= P**

_{1}**V**

_{2}**/T**

_{2}

_{2}P

**× 22.4/273 = 2P**

_{1}**V**

_{2}**/546**

_{2}V

**= 22.4 litre**

_{2}**Solution 22:**

(a) The molecular mass of (Mg(NO

**)**

_{3}**.6H**

_{2}**O = 256.4 g**

_{2}% of Oxygen = 12 × 16/256 = 75%

**B**

_{2}**O**

_{4}**.10H**

_{7}**O = 382 g**

_{2}% of B = 4 × 11/382 = 11.5%

**Solution 23:**

(V×760)/273 = (360 × 380)/360

V = (360 × 380 × 273)/(760 × 360) = 136.5cm

^{3}136.5cm

**of the gas weigh = 0.546**

^{3}22400 cm

**of the gas weight = (0.546 × 22400)/136.5 = 89.6 a.m.u**

^{3}Relative molecular mass = 89.6 a.m.u

**Solution 24:**

(a) 252 g of solid ammonium dichromate decomposes to give 152 g of solid chromium oxide, so the loss in mass in terms of solid formed = 100 g

Now, if 63 g ammonium dichromate is decomposed, the loss in mass would be = (100 × 63)/252 = 25 g

**O**

_{2}**= 152 g**

_{3}So, 63 g ammonium dichromate will produce = (63 × 152/252) = 38 g

**Solution 25 :**

**gives = 2 × 22.4 litres volume**

_{2}So, 12.8 g of SO

**gives = (2 × 22.4 × 12.8)/128 = 4.48 litre volume**

_{2}Or one can say 4.48 litres of hydrogen sulphide.

2 x 22.4 litre H

**S requires oxygen = 3 × 22.4 litre**

_{2}So, 4.48 litres H

**S will require = 6.72 litres of oxygen**

_{2}**Solution 26**:

From equation, 2NH

**+ 2O**

_{3}**→ 2NO + 3H**

_{2}**O**

_{2}When 60 g NO is formed, the mass of steam produced = 54g

So, 1.5 g NO is formed, the mass of steam produced = (54 × 1.5/60) = 1.35 g

**Solution 27:**

In 1 hectare of soil, N

**removed = 20 kg**

_{2}So, in 10 hectare N

**removed = 200 kg**

_{2}The molecular mass of Ca(NO

**)**

_{3}**= 164**

_{2}Now, 28 g N

**present in fertilizer = 164 g Ca(NO**

_{2}**)**

_{3}

_{2}So, 200000 g of N

**is present in = (164 × 200000)/28**

_{2}= 1171.42 kg

**Solution 28:**

(a) 1 mole of phosphorus atom = 31 g of phosphorus

31 g of P = 1 mole of P

6.2g of P = (6.2 × 1)/31 = 0.2 mole of P

(b) 31 g P reacts with HNO

**= 315 g**

_{3}So, 6.2 g P will react with HNO

**= (315 × 6.2/31) = 63 g**

_{3}Moles of steam formed from 6.2 g phosphorus = 1mol/3196.2 = 0.2 mol

Volume of steam produced at STP = (0.2 × 22.4l)/MOL = 4.48 litre

Since the pressure (760mm) remains constant , but the temperature (273 + 273) = 546 is double, the volume of the steam also gets doubled

So,Volume of steam produced at 760mm Hg and 273˚C = 4.48 × 2 = 8.96litre

**Solution 29:**

(a) 1 mole of gas occupies volume = 22.4 litre

(b) 112cm

**of gaseous fluoride has mass = 0.63 g**

^{3}So, 22400cm

**will have mass = 0.63 × 22400/112 = 126 g**

^{3}The molecular mass = At mass P + At mass of F

126 = 31 + At. Mass of F

So, At. Mass of F = 95 g

But, at. mass of F = 19 so 95/19 = 5

Hence, there are 5 atoms of F so the molecular formula = PF

_{5}**Solution 30:**

Na

**CO**

_{2}**.10H**

_{3}**O → Na**

_{2}**CO**

_{2}**+ 10H**

_{3}**O**

_{2}286 g 106g

So, for 57.2 g Na

**CO**

_{2}**.10H**

_{3}**O = 106 ×× 57.2/286 = 21.2 g Na**

_{2}**CO**

_{2}

_{3}**Solution 31:**

(a) The molecular mass of Ca(H

**PO**

_{2}**)**

_{4}**= 234**

_{2}The % of P = (2 × 31/234) = 26.49%

(b) Simple ratio of M = 34.5/56 = 0.616 = 1

Simple ratio of CI = 65.5/35.5 = 1.845 = 3

Empirical formula = MCI

_{3}Empirical formula mass = 162.5, Molecular mass = 2 ×V.D. = 325

So, n = 2

So, molecular formula = M

**Cl**

_{2}

_{6}**Solution 32:**

**V**

**/V**

_{1}**= n**

_{2}**/n**

_{1}

_{2}So, no. of moles of Cl = x/2 (since V is directly proportional to n)

No. of moles of NH

**= x**

_{3}No. of moles of SO

**= x/4**

_{2}This is because of Avogadro's law which states Equal volumes of all gases, under similar conditions of temperature and pressure, contain an equal number of molecules.

So, 20 litre nitrogen contains x molecules

So, 10 litre of chlorine will contain = x × 10/20 = x/2 mols.

And, 20 litre of ammonia will also contain = x molecules

And, 5 litre of sulphur dioxide will contain = x × 5/20 = x/4 mols.

**Solution 33:**

2 × 22400 litre steam is produced by N

**O = 4 × 22400 cm**

_{2}

^{3}So, 150 cm

**steam will be produced by = (4 × 22400 ×150)/(2 ×22400)**

^{3}= 300 cm

**N**

^{3}**O**

_{2}**Solution 34:**

**(a) Volume of O**

**= V**

_{2}Since O

**and N**

_{2}**have same no. of molecules = x**

_{2}so, the volume of N

**= V**

_{2}

_{2}So, 8 g oxygen is contained in = (44 × 8)/32 = 11 g

(d) Avogadro’s law is used in the above questions.

**Solution 35:**

(a) 444 g is the molecular formula of (NH

**)**

_{4}**PtCl**

_{2}

_{6}% of Pt = (195/444) × 100 = 43.91% or 44%

(b) simple ratio of Na = 42.1/23 = 1.83 = 3

simple ratio of P = 18.9/31 = 0.609 = 1

simple ratio of O = 39/16 = 2.43 = 4

So, the empirical formula is Na

**PO**

_{3}

_{4}**Solution 36:**

22.4 litres of methane requires oxygen = 44.8 litres O

_{2}From equation,

44.8 litres hydrogen requires oxygen = 22.4 litres O

_{2}So, 11.2 litres will require = (22.4 × 11.2)/44.8 = 5.6 litres

Total volume = 44.8 + 5.6 = 50.4 litres

**Solution 37:**

According to Avogadro's law:

Equal volumes of all gases, under similar conditions of temperature and pressure, contain an equal number of molecules.

So, 1 mole of each gas contains = 6.02 ×10

**molecules**

^{23}Mol. Mass of H

**(2), O**

_{2}**(32), CO**

_{2}**(44), SO**

_{2}**(64),Cl**

_{2}**(71)**

_{2}

^{23}**)**

^{23}**= 4 × 6.02 ×10**

^{ }**= 4M molecules**

^{23}**= M/4**

^{23}**= 2M/11**

^{23}

^{23}So, 8g of sulphur dioxide molecules = 8/64 × 6.02 ×10

**= M/8**

^{23}

^{23}So, 8g of chlorine molecules = 8/72 × 6.02 × 10

**= 8M/71**

^{23}Since 8M/71<M/8<2M/11<M/4<4M

Thus Cl

**<SO**

_{2}**<CO**

_{2}**<O**

_{2}**<H**

_{2}

_{2}(i) Least number of molecules in Cl

_{2}(ii) Most number of molecules in H

_{2}**Solution 38:**

**Na**

**SO**

_{2}**+ BaCl**

_{4}**→ BaSO**

_{2}**+ 2NaCI**

_{4}Molecular mass of BaSO

**= 233 g**

_{4}Now, 233 g of BaSO

**is produced by Na**

_{4}**SO**

_{2}**= 142 g**

_{4}So, 6.99 g BaSO

**will be produced by = (6.99 × 142/233) = 4.26**

_{4}The percentage of Na

**SO**

_{2}**in original mixture = (4.26 × 100/10) = 42.6%**

_{4}**Solution 39:**

**(a) 1 litre of oxygen has mass = 1.32 g**

So, 24 litres (molar vol. at room temp.) will have mass = 1.32 ×24 = 31.6 or 32g

**→ K**

_{4}**MnO**

_{2}**+ MnO**

_{4}**+ O**

_{2}

_{2}316 g of KMnO

**gives oxygen = 24 litres**

_{4}So, 15.8 g of KMnO

**will give = (24 × 316/15.8) = 1.2 litres**

_{4}**Solution 40:**

**(a) (i) The no. of moles of SO**

**= 3.2/64 = 0.05 moles**

_{2}**, no. of molecules present = 6.02 ×10**

_{2}

^{23}**× 0.05 = 3.0 ×10**

^{23}

^{22}**= 22.4 dm**

_{2}

^{3}3.2 g of SO

**will be occupied by volume = (22.4 × 3.2/64) = 1.12 dm**

_{2}

^{3}Gram atoms of Cl = 4.26/35.5 = 0.12 = 4

So, the empirical formula = PbCl

_{4}**Solution 41:**

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

V

**/V**

_{1}**= n**

_{2}**/n**

_{1}

_{2}V

**/V**

_{1}**= n**

_{2}**/2n**

_{1}

_{1}So, V

**= 2V**

_{2}

_{1}**, so the number of molecules = 6 ×10**

^{3}**because according to mole concept 22.4 litre volume at STP has = 6 ×10**

^{23}**molecules**

^{23}so, mass of N

**O = 1 44 = 44 g**

_{2}

**Solution 42:**

(a) NaCl + NH

**+ CO**

_{3}**+ H**

_{2}**O → NaHCO**

_{2}**+ NH**

_{3}**CI**

_{4}2NaHCO

**→ Na**

_{3}**CO**

_{2}**+ H**

_{3 }**O + CO**

_{2}

_{2}From equation:

106 g of Na

**CO**

_{2}**is produced by = 168g of NaHCO**

_{3}

_{3}So, 21.2 g of Na

**CO**

_{2}**will be produced by = (168 × 21.2)/106**

_{3}= 33.6 g of NaHCO

_{3}**, required volume of CO**

_{3}**= 22.4 litre**

_{2}So, for 33.6 g of NaHCO

**, required volume of CO**

_{3}**= 22.4 × 33.6/84**

_{2}= 8.96 litre

**Solution 43:**

(a)

44.8 litres of water produced by = 22.4 litres of NH

**NO**

_{4}

_{3}So, 8.96 litres will be produced by = (22.4 × 8.96)/44.8

= 4.48 litres of NH

**NO**

_{4}

_{3}So, 4.48 litres of N

**O is produced.**

_{2}**O is produced by = 80 g of NH**

_{2}**NO**

_{4}

_{3}So, 8.96 litres H

**O will be produced by = (80 × 8.96/44.8)**

_{2}= 16g NH

**NO**

_{4}

_{3}**NO**

_{4}**= (3 × 16)/80 = 60%**

_{3}**Solution 44:**

(a)

Element |
% |
At. Mass |
Atomic ratio |
Simplest ratio |

K |
47.9 |
39 |
1.22 |
2 |

Be |
5.5 |
9 |
0.6 |
1 |

F |
46.6 |
19 |
2.45 |
4 |

**BeF**

_{2}

_{4}3 × 80 g of CuO reacts with = 2 × 22.4 litre of NH

_{3}so, 120 g of CuO will react with = (2 × 22.4 × 120)/(80 × 3) = 22.4 litres

**Solution 45:**

**(a) The molecular mass of ethylene(C**

**H**

_{2}**) is 28 g**

_{4}No. of moles = 1.4/28 = 0.05 moles

No. of molecules = (6.023 × 10

**× 0.05)= 3 × 10**

^{3}**molecules**

^{22}Volume = 22.4 x 0.05 = 1.12 litres

S0, V.D = 28/2 = 14

**Solution 46:**

**(a) Molecular mass of Na**

**AlF**

_{3}**= 210**

_{6}So, Percentage of Na = (3 × 23 × 100)/210 = 32.85%

1 mole of O

**has volume = 22400 ml**

_{2}Volume of oxygen used by 2 x 22400 ml CO = 22400 ml

So, Vol. of O

**used by 560 ml CO = (22400 × 560)/(2 × 22400) = 280 ml**

_{2}So, Volume of CO

**formed is 560 ml.**

_{2}**Solution 47:**

(a) Mass of gas X = 10g

Mass of hydrogen gas = 2

Relative vapour density

= Mass of volume of gas X under similar conditions/ Mass of volume of hydrogen gas under similar conditions = 10/2 = 5

Relative molecular mass of the gas = 2 × relative vapour

density = 2 × 5 = 10

**H**

_{2}**(g) + 5O**

_{2 }**(g) → 4CO**

_{2}**(g) + 2H**

_{2}**O(g)**

_{2}According to Gay-Lussac's law,

2 volume of acetylene requires 5 volume of oxygen to burn it

∴ 1 volume of acetylene requires 2.5 volume of oxygen to burn it

∴ 200cm

**requires 2.5 ×200 = 500 cm**

^{3}**of oxygen**

^{3}2 volume of acetylene on combustion gives 4CO

_{2}∴ 1 volume of acetylene on combustion gives 2CO

_{2}∴ 200cc of acetylene on combustion will give 200 × 2 = 400cc of CO

_{2}∴ Nitrogen = 100 - 12.5 = 87.5

The Empirical formula of the compound is NH

_{2}Empirical formula weight = 14 + 2 = 16

Relative molecular mass = 37

N = Relative molecular mass/Empirical Weight = 37/16 = 2.3≈2

Molecular formula = n ×empirical formula = 2 × NH

**= N**

_{2 }**H**

_{2}

_{4}

^{24}Avogadro's number = 6 ×10

^{23}1. Mass of nitrogen in a cylinder = (24 × 10

**× 28)/(6 ×10**

^{24}**) = 1120g**

^{23}2. Volume of nitrogen at stp

Volume of 28 g of N

**= 22.4dm**

_{2}

^{3}Volume of 1120g of N

**= (1120 × 22.4)/28 dm**

_{2}**= 896 dm**

^{3 }

^{3}

**Solution 48:**

**a. (i) 10 litres of LPG contains**

Propane = (60/100) × 10 = 6lites

Butane = (40/100) × 10 = 4litres

CO

**= 18 + 16 = 34l**

_{2}**(NO**

_{4}**) = 80**

_{3}H = 1, N = 14, O = 16

**% of Nitrogen**

As 80 g of NH

**(NO**

_{4}**) contains 28 g of nitrogen**

_{3}∴ 100 g of of NH

**(NO**

_{4}**) will contain (28 × 100)/80 = 35%**

_{3}**% of Oxygen**

As, 80 g of NH

**(NO**

_{4}**) contains 48 g of oxygen**

_{3}∴ 100 g of of NH

**(NO**

_{4}**) will contain (100 × 48)/80 = 60%**

_{3}b. (i) Equation for reaction of calcium carbonate with dilute hydrochloric acid:

CaCO

**+ 2HCl → CaCl**

_{3}**+ H**

_{2}**O + CO**

_{2}**↑**

_{2}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

**+ 2HCl → CaCl**

_{3}**+ H**

_{2}**O + CO**

_{2}**↑**

_{2}As, 100g of calcium carbonate gives 22.4dm3 of CO

_{2}∴ 450 g of calcium carbonate will give 100 = 100.8 L

Relative molecular mass of calcium chloride = 111

As 100 g of calcium carbonate gives 111g of calcium chloride

∴ 450 g of calcium carbonate will give (450 × 111/100) = 499.5 g

The molecular mass of calcium carbonate = 100

As 100 g of calcium carbonate gives (2 ×36.5) = 73g of HCI

∴ 450 g of calcium carbonate will give (450 × 73/100) = 328.5g

Number of moles of HCl = Weight of HCl/Molecular weight of HCI = 328.5/36.5 = 9 moles

**Solution 49:**

a. (i) Atomic mass: S = 32 and O = 16

Molecular mass of SO

**= 32 + (2 ×16) = 64g**

_{2}As 64 g of SO

**= 22.4dm**

_{2}

^{3}Then, 320 g of SO

**= (320 × 22.4)/64**

_{2}= 112 L

**H**

_{3}**+ 5O**

_{8}**→ 3CO**

_{2}**+ 4H**

_{2}**O**

_{2}Molar mass of propane = 44

44 g of propane requires 5 × 22.4 litres of oxygen at STP.

8.8 g of propane requires (5 × 22.4 × 8.8)/44 = 22.4 litres

Empirical formula = CH

**Br**

_{2}n(Empirical formula mass of CH

**Br) = Molecular mass (2 × VD)**

_{2}n(12 + 2 + 80) = 94 ×2

n = 2

Molecular formula = Empirical formula ×2

= (CH

**Br) × 2**

_{2}= C

**H**

_{2}**Br**

_{4}

_{2}**atoms of sulphur**

^{22}6.022 × 10

**atoms of sulphur will have mass = 32 g**

^{23}**atoms of sulphur will have mass = (32 ×10**

^{22}**)/(6.022 ×10**

^{22}**)= 0.533 g**

^{23}1 mole of carbon dioxide will have mass = 44 g

0.1 mole of carbon dioxide will have mass = 4.4 g

**Solution 50:**

a. P + 5HNO

_{3}(conc.) → H

**PO**

_{3}**+ H**

_{4}**O + 5NO**

_{2}

_{2}(i) Number of moles of phosphorus taken = 9.3/31 = 0.3 mol

So, 0.3 mole of phosphorus gives (0.3 × 98) gm of phosphoric acid

= 29.4 gm of phosphoric acid

**gas at STP.**

_{2}So, 0.3 mole of phosphorus gives (112 × 0.3) L of

NO

**gas at STP.**

_{2}= 33.6 L of NO

**gas at STP**

_{2}N

**(g) + 3H**

_{2}**(g) → 2NH**

_{2}**(g)**

_{3}3 volumes of hydrogen produce 2 volumes of ammonia

67.2 litres of hydrogen produce (2 × 67.2)/3 = 44.8 L

3 volumes of hydrogen combine with 1 volume of ammonia.

67.2 litres of hydrogen combine with (1 × 67.2)/3 = 22.4L

Nitrogen left = 44.8 - 22.4 = 22.4 litres

(ii) 5.6 dm

**of gas weighs 12 g**

^{3}1 dm

**of gas weighs = (12/56) gm**

^{3}22.4 dm

**of gas weighs = (12/56 × 22.2) gm = 48g**

^{3}Therefore, the relative molecular mass of gas = 48 gm.

(iii) Molar mass of Mg (NO

**)**

_{3}**.6H**

_{2}**O**

_{2}= 24 × (14 × 2) + (16 × 12) + (1 × 12) = 256 g

24 × 100 = 9.37%

Mass percent of magnesium = (24 × 100)/256 = 9.37%

**Solution 51:**

a.

(i) 2 vols. of butane requires O

**= 13 vols**

_{2}90 dm

**of butane will require O**

^{3}**= 13/2 × 90 = 585 dm**

_{2}

^{3}So, molecular mass of gas = 2 × 8 = 16 g

As we know, molecular mass or molar mass occupies 22.4 litres.

That is,

16 g of gas occupies volume = 22.4 litres

So, 24 g of gas will occupy a volume

22.4 × 24 = 33.6 litres

So, molecules of nitrogen gas present in the same vessel = X

b.

(i) 3 vols. of oxygen require KClO

**= 2 vols.**

_{3}So, 1 vol. of oxygen will require KClO

**= 2/3 vols**

_{3}So, 6.72 litres of oxygen will require KCIO

_{3}So, 1 vol. of oxygen will require KClO

**= 2/3vols**

_{3}So, 6.72 litres of oxygen will require KClO

_{3}2/3 x 6.72 = 4.48 litres

22.4 litres of KClO

**has mass = 122.5 g**

_{3}So, 4.48 litres of KÄŒIO

**will have mass**

_{3}= (122.5/22.4 x 4.48) = 24.5g

(ii) 22.4 litres of oxygen = 1 mole

So, 6.72 litres of oxygen = 6.72/22.4 = 0.3 moles

No. of molecules present in 1 mole of O

**= 6.023 ×10**

_{2}

^{23}So, no. of molecules present in 0.3 mole of O

_{2}= 6.023 × 10

**× 0.3**

^{23}= 1.806 × 10

^{23}**at STP = 22.4 litres**

_{2}So, volume occupied by 0.01 mole of CO

**at STP = 22.4 × 0.01= 0.224 litres**

_{2}**Solution 52:**

a. (i) 2C

**H**

_{2}**+ 5O**

_{2}**→ 4CO**

_{2}**+ 2H**

_{2}**O**

_{2}2 moles of C

**H**

_{2}**= 4moles of CO**

_{2}

_{2}x dm

**of C**

^{3}**H**

_{2}**= 8.4 dm**

_{2}**of CO**

^{3}

_{2}x = (2 × 8.4)/4

= 4.2 dm

**of C**

^{3}**H**

_{2}

_{2}**Y**

_{2}Atomic weight (X) = 10

Atomic weight (Y) = 5

Empirical formula weight = (2 × 10) + 5 = 25

n = (Molecular weight/Emperical formula weight)

= (2 × V.D)/Emperical formula weight

= (2 × 25)/25 = 2

So, molecular formula = X

**Y**

_{2}

_{2}

Molecular weight of ammonia = 17g/mole

68g of ammonia gas at STP =?

1 mole = 22.4 dm

^{3}∴ 4 mole = 22.4 × 4 = 89.6 dm

^{3}**molecules**

^{23}4 moles = 24.092 ×10

**molecules**

^{23 }