**Selina ICSE Solutions for Class 10 Chemistry – Mole Concept and Stoichiometry**

Download Formulae Handbook For ICSE Class 9 and 10

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

**Exercise 5(A)**

**Solution 1.**

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

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

**Solution 2.**

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

b) N_{2}means 1 molecule of nitrogen and 2N means two atoms of nitrogen.

N_{2} can exist independently but 2N cannot exist independently.

**Solution 3.**

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

Now volume of hydrogen gas =volume of helium gas

n molecules of hydrogen =n molecules of helium gas

nH_{2}=nHe

1 mol. of hydrogen has 2 atoms of hydrogen and I molecule of helium has 1 atom of helium

Therefore 2H=He

Therefore atoms in hydrogen is double the atoms of helium.

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

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

**Solution 4.**

2H_{2 }+ O_{2} → 2H_{2}O

2 V 1V 2V

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

so 200cm^{3} of Hydrogen reacts with = 200/2= 100 cm^{3
}Hence, the unreacted oxygen is 150 – 100 = 50cm^{3 }of oxygen.

**Solution 5.**

**Solution 6.**

**Solution 7.**

**Solution 8.**

**Solution 9.**

**Solution 10.**

**Solution 11.**

C_{3}H_{8 }+ 5O_{2} → 3CO_{2} + 4H_{2}O

1 V 5 V 3 V

From equation, 5 V of O_{2} required = 1V of propane

so, 100 cm^{3} of O_{2 }will require = 20 cm^{3} of propane

**Solution 12.**

**Solution 13.**

2CO + O_{2} → 2CO_{2
}2 V 1 V 2 V

2 V of CO requires = 1V of O_{2
}so, 100 litres of CO requires = 50 litre of O_{2}

**Solution 14.**

**Solution 15.**

H_{2} + Cl_{2 }→ 2HCl

1V 1V 2 V

Since 1 V hydrogen requires 1 V of oxygen and 4cm^{3} of H_{2} remained behind so the mixture had com”>16 cm^{3} hydrogen and 16 cm^{3 }chlorine.

Therefore Resulting mixture is H_{2} =4cm^{3},HCl=32cm^{3}

**Solution 16.**

CH_{4} + 2O_{2 }→ CO_{2} + 2H_{2}O

1 V 2 V 1 V

2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}O

2 V 5 V 4 V

From the equations, we can see that

1V CH_{4} requires oxygen = 2 V O_{2
}So, 10cm^{3} CH_{4} will require =20 cm^{3} O_{2
}Similarly 2 V C_{2}H_{2} requires = 5 V O_{2
}So, 10 cm^{3} C_{2}H_{2 }will require_{ }= 25 cm^{3} O_{2
}Now, 20 V O_{2} will be present in 100 V air and 25 V O_{2} will be present in 125 V air ,so the volume of air required is 225cm^{3}

**Solution 17.**

**Solution 18.**

**Solution 19.**

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

Since the unchanged or remaining O_{2} is 58 cc so, used oxygen 106 – 58 = 48cc

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

CH_{4} + 2O_{2} → CO_{2} + 2H_{2}O

1 V 2 V

24 cc 48 cc

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

**Solution 20.**

According to Avogadro’s law, equal volumes of gases contain equal no. of molecules under similar conditions of temperature and pressure. This means more volume will contain more molecules and least volume will contain least molecules.

So,

(a) 5 litres of hydrogen has greatest no. of molecules with the maximum volume.

(b) 1 litre of SO_{2} contains the least number of molecules since it has the smallest volume.

**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^{3 }at STP

**Solution 2.**

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

(b) Molar volume is the volume occupied by one mole of the gas at STP. It is equal to 22.4 dm^{3}.

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

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

(e) The number of atoms present in 12g (gram atomic mass) of C-12 isotope, i.e. 6.023 x10^{23} atoms.

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

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

**Solution 3.**

(a) Applications of Avogadro’s Law :

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

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

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

H_{2} + Cl_{2} → 2HCl

1V 1V 2V (By Gay-Lussacs law)

n molecules n molecules 2n molecules (By Avogadros law)

**Solution 4.**

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

(b) KClO_{3} = (K)39 + (Cl)35.5 + (3O)48 = 122.5

(c) (Cu)63.5 + (S)32 + (4O)64 + (5H_{2}O)5 x 18 = 249.5

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

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

(f) (C)12 + (H)1+ (3Cl)3 x 35.5 = 119.5

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

**Solution 5.**

(a) No. of molecules in 73 g HCl = 6.023 x10^{23 }x 73/36.5(mol. mass of HCl)

= 12.04 x 10^{23}

(b) Weight of 0.5 mole of O_{2} is = 32(mol. Mass of O_{2}) x 0.5=16 g

(c) No. of molecules in 1.8 g H_{2}O = 6.023 x 10^{23 }x 1.8/18

= 6.023 x 10^{22}

(d) No. of moles in 10g of CaCO_{3} = 10/100(mol. Mass CaCO_{3})

= 0.1 mole

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

(f) No. of molecules in 3.2 g of SO_{2} = 6.023 x 10^{23} x 3.2/64

= 3.023 x 10^{22}

**Solution 6.**

Molecular mass of H_{2}O is 18, CO_{2} is 44, NH_{3} is 17 and CO is 28

So, the weight of 1 mole of CO_{2} is more than the other three.

**Solution 7.**

4g of NH_{3} having minimum molecular mass contain maximum molecules.

**Solution 8.
**

a) No. of particles in s1 mole = 6.023 x 10^{23
}So, particles in 0.1 mole = 6.023 x 10 ^{23} x 0.1 = 6.023 x 10^{22}

b) 1 mole of H_{2}SO_{4} contains =2 x 6.023 x 10^{23
}So, 0.1 mole of H_{2}SO_{4} contains =2 x 6.023 x 10^{23} x0.1

= 1.2×10^{23 }atoms of hydrogen

c) 111g CaCl_{2} contains = 6.023 x 10^{23 }molecules

So, 1000 g contains = 5.42 x 10^{24 }molecules

**Solution 9.**

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

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

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

= 3.65 g

(c) 0.2 mole of H_{2}O has mass = 0.2 x 18 = 3.6 g

(d) 0.1 mole of CO_{2} has mass = 0.1 x 44 = 4.4 g

**Solution 10.**

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

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

= 48g(molar mass)

(b) 1 mole of SO_{2} has volume = 22.4 litres

So, 2 moles will have = 22.4 x 2 = 44.8 litre

**Solution 11.**

(a) 1 mole of CO_{2} contains O_{2} = 32g

So, CO_{2 }having 8 gm of O_{2 }has no. of moles = 8/32 = 0.25 moles

(b) 16 g of methane has no. of moles = 1

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

**Solution 12.**

(a) 6.023 x 10 ^{23 }atoms of oxygen has mass = 16 g

So, 1 atom has mass = 16/6.023 x 10^{23 }= 2.656 x 10^{-23 }g

(b) 1 atom of Hydrogen has mass = 1/6.023 x 10^{23 }= 1.666 x 10^{-24
}(c) 1 molecule of NH_{3} has mass = 17/6.023 x10^{23 }= 2.82 x 10^{-23 }g

(d) 1 atom of silver has mass = 108/6.023 x 10^{23 }=1.701 x 10^{-22
}(e) 1 molecule of O_{2} has mass = 32/6.023 x 10^{23 }= 5.314 x 10^{-23 }g

(f) 0.25 gram atom of calcium has mass = 0.25 x 40 = 10g

**Solution 13.**

(a) 0.1 mole of CaCO_{3} has mass =100(molar mass) x 0.1=10 g

(b) 0.1 mole of Na_{2}SO_{4}.10H_{2}O has mass = 322 x 0.1 = 32.2 g

(c) 0.1 mole of CaCl_{2} has mass = 111 x 0.1 = 11.1g

(d) 0.1 mole of Mg has mass = 24 x 0.1 = 2.4 g

**Solution 14.**

1molecule of Na_{2}CO_{3}.10H_{2}O contains oxygen atoms = 13

So, 6.023 x10^{23 }molecules (1mole) has atoms=13 x 6.023 x 10^{23
}So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10^{23} =7.8 x 10^{23}

**Solution 15.**

3.2 g of S has number of atoms = 6.023 x10^{23} x 3.2 /32

= 0.6023 x 10^{23
}So, 0.6023 x 10^{23} atoms of Ca has mass=40 x0.6023×10^{23}/6.023 x 10^{23
}= 4g

**Solution 16.**

(a) No. of atoms = 52 x 6.023 x10^{23} = 3.131 x 10^{25
}(b) 4 amu = 1 atom of He

so, 52 amu = 13 atoms of He

(c) 4 g of He has atoms = 6.023 x10^{23
}So, 52 g will have = 6.023 x 10^{23} x 52/4 = 7.828 x10^{24} atoms

**Solution 17.**

Molecular mass of Na_{2}CO_{3} = 106 g

106 g has 2 x 6.023 x10^{23} atoms of Na

So, 5.3g will have = 2 x 6.023 x10^{23}x 5.3/106=6.022 x10^{22 }atoms

Number of atoms of C = 6.023 x10^{23 }x 5.3/106 = 3.01 x 10^{22 }atoms

And atoms of O = 3 x 6.023 x 10^{23 }x 5.3/106= 9.03 x10^{22 }atoms

**Solution 18.**

(a) 60 g urea has mass of nitrogen(N_{2}) = 28 g

So, 5000 g urea will have mass = 28 x 5000/60 = 2.33 kg

(b) 64 g has volume = 22.4 litre

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

**Solution 19.**

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

(b) Vapour density of Chlorine atom is 35.5.

**Solution 20.**

22400 cm^{3} of CO has mass = 28 g

So, 56 cm^{3} will have mass = 56 x 28/22400 = 0.07 g

**Solution 21.**

18 g of water has number of molecules = 6.023 x 10^{23
}So, 0.09 g of water will have no. of molecules = 6.023 x 10^{23} x 0.09/18 = 3.01 x 10^{21 }molecules

**Solution 22.**

(a) No. of moles in 256 g S_{8} = 1 mole

So, no. of moles in 5.12 g = 5.12/256 = 0.02 moles

(b) No. of molecules = 0.02 x 6.023 x 10^{23} = 1.2 x 10^{22 }molecules

No. of atoms in 1 molecule of S_{ }= 8

So, no. of atoms in 1.2 x 10^{22 }molecules = 1.2 x 10^{22 }x 8

= 9.635x 10^{22 }molecules

**Solution 23.**

Atomic mass of phosphorus P = 30.97 g

Hence, molar mass of P_{4} = 123.88 g

If phosphorus is considered as P_{4} molecules,

then 1 mole P_{4 }≡ 123.88 g

Therefore, 100 g of P_{4 }= 0.807 g

**Solution 24.**

**Solution 25.**

No. of atoms in 12 g C = 6.023 x10^{23
}So, no. of carbon atoms in 10^{-12} g = 10^{-12} x 6.023 x10^{23}/12

= 5.019 x 10^{10 }atoms

**Solution 26.**

Given:

P= 1140 mm Hg

Density = D = 2.4 g / L

T = 273 ^{0}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 liter of unknown gas at standard temperature 273 K.

V_{1}= 1 L T_{1} = 546 K

V_{2}=? T_{2} = 273 K

V_{1}/T_{1} = V_{2}/ T_{2
}V_{2} = (V_{1} x T_{2})/T_{1
} = (1 L x 273 K)/546 K

= 0.5 L

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

Apply Boyle’s law.

P _{1} = 1140 mm Hg V_{1} = 0.5 L

P_{2} = 760 mm Hg V_{2} = ?

P_{1} x V_{1 }= P_{2} x V_{2
}V_{2} = (P_{1} x V_{1})/P_{2
} = (1140 mm Hg x 0.5 L)/760 mm Hg

= 0.75 L

Now, 22.4 L is the volume of 1 mole of any gas at STP, then 0.75 L is the volume of X moles at STP

X moles = 0.75 L / 22.4 L

= 0.0335 moles

The original mass is 2.4 g

n = m / M

0.0335 moles = 2.4 g / M

M = 2.4 g / 0.0335 moles

M= 71.6 g / mole

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

**Solution 27.**

1000 g of sugar costs = Rs. 40

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

**Solution 28.**

**Solution 29.**

40 g of NaOH contains 6.023 x 10^{23 }molecules

So, 4 g of NaOH contains = 6.02 x10^{23} x 4/40

= 6.02 x10^{22} molecules

**Solution 30.**

The number of molecules in 18 g of ammonia= 6.02 x10^{23
}So, no. of molecules in 4.25 g of ammonia = 6.02 x 10^{23 }x 4.25/18

= 1.5 x 10^{23}

**Solution 31.**

(a) One mole of chlorine contains 6.023 x 10^{23} atoms of chlorine.

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

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

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

**Exercise 5(C)**

**Solution 1.**

Information conveyed by H_{2}O

- That H
_{2}O contains 2 volumes of hydrogen and 1 volume of oxygen. - That ratio by weight of hydrogen and oxygen is 1:8.
- That molecular weight of H
_{2}O is 18g.

**Solution 2.**

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

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

**Solution 3.**

(a) CH (b) CH_{2}O (c) CH (d) CH_{2}O

**Solution 4.**

**Solution 5.**

**Solution 6.**

Molecular mass of KClO_{3 }= 122.5 g

% of K = 39 /122.5 = 31.8%

% of Cl = 35.5/122.5 = 28.98%

% of O = 3 x 16/122.5 = 39.18%

**Solution 7.**

**Solution 8.**

**Solution 9.**

**Solution 10.**

**Solution 11.**

**Solution 12.**

**Solution 13.**

**Solution 14.**

**Solution 15.**

**Solution 16.**

**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_{3 }N_{2}

**Solution 18.**

Barium chloride = BaCl_{2}.*x *H_{2}O

Ba + 2Cl + x[H_{2} + O]

= 137+ 235.5 + *x [2+16]
*= [208 + 18

*x*] contains water = 14.8% water in BaCl

_{2}.

*x*H

_{2}O

= [208 + 18

*x]*14.8/100 = 18

*x*

= [104 + 9

*x*] 2148=18000

*x*

= [104+9

*x*] 37=250

*x*

= 3848 + 333

*x =2250x*

1917

*x =*3848

*x =*2molecules of water

**Solution 19.**

Molar mass of urea; CON_{2}H_{4 }= 60 g

So, % of Nitrogen = 28 × 100/60 = 46.66%

**Solution 20.**

Element % At. mass Atomic ratio Simple ratio

C 42.1 12 3.5 1

H 6.48 1 6.48 2

O 51.42 16 3.2 1

The empirical formula is CH_{2}O

Since the compound has 12 atoms of carbon, so the formula is

C_{12} H_{24} O_{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, molecular formula is A_{2}B_{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_{6}B_{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

Empirical formula of the given compound =ZnSH_{14}O_{11
}Empiricala formula mass = 65.37+32+141+11+16=287.37

Molecular mass = 287

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

Molecular formula = ZnSO_{11}H_{14
}= ZnSO_{4}.7H_{2}O

**Exercise 5(D)**

**Solution 1.**

**Solution 2.**

**Solution 3.**

**Solution 4.**

**Solution 5.**

Molecular mass of KNO_{3 }= 101 g

63 g of HNO_{3} is formed by = 101 g of KNO_{3
}So, 126000 g of HNO_{3} is formed by = 126000 x 101/63 = 202 kg

Similarly,126 g of HNO_{3} is formed by 170 kg of NaNO_{3
}So, smaller mass of NaNO_{3 }is required.

**Solution 6.**

**Solution 7.**

**Solution 8.**

**Solution 9.**

**Solution 10.**

**Solution 11.**

**Solution 12.**

**Miscellaneous Exercise**

**Solution 1.**

**Solution 2.**

**Solution 3.**

**Solution 4.**

**Solution 5.**

**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.**

**Solution 8.**

**Solution 9.**

Mass of X in the given compound =24g

Mass of oxygen in the given compound =64g

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

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

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

Element % At. Mass Atomic ratio Simplest ratio

X 27.3 12 27.3/12=2.27 1

O 72.7 16 72.2/16=4.54 2

So simplest formula = XO_{2}

**Solution 10.**

**Solution 11.**

**Solution 12.**

**Solution 13.**

(a) Number of molecules in 100cm^{3} of oxygen=Y

According to Avogadros law, Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.Therefore ,number of molecules in 100 cm^{3} of nitrogen under the same conditions of temperature and pressure = Y

So, number of molecules in 50 cm^{3} of nitrogen under the same conditions of temperature and pressure =Y/100 50=Y/2

(b) (i) Empirical formula is the formula which tells about the simplest ratio of combining capacity of elements present in a compound.

(ii) The empirical formula is CH_{3
}(iii) The empirical formula mass for CH_{2}O = 30

V.D = 30

Molecular formula mass = V.D 2 = 60

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

So, molecular formula = (CH_{2}O)_{2} = C_{2}H_{4}O_{2}

**Solution 14.**

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

**Solution 15.**

**Solution 16.**

**Solution 17.**

**Solution 18.**

So, mass of CO_{2} = 22 kg

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

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

**Solution 19.**

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

(b) Since PV=constant so, if pressure is doubled; the volume will become half i.e. 11.2 litres.

(c) V_{1}/V_{2} = T_{1}/T_{2
}22.4/V_{2} =273/546

V_{2} = 44.8 litres

(d) Mass of 1 mole Cl_{2} gas =35.5 x 2 =71 g

**Solution 20.**

**Solution 21.**

**Solution 22.**

(a) The molecular mass of (Mg(NO_{3})_{2}.6H_{2}O = 256.4 g

% of Oxygen = 12 x 16/256

= 75%

(b) The molecular mass of boron in Na_{2}B_{4}O_{7}.10H_{2}O = 382 g

% of B = 4 x 11/382 = 11.5%

**Solution 23.**

**Solution 24.**

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

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

(b) If 252 g of ammonium dichromate produces Cr_{2}O_{3 }= 152 g

So, 63 g ammonium dichromate will produce = 63 x 152/252

= 38 g

**Solution 25.**

**Solution 26.**

**Solution 27.**

**Solution 28.**

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^{0}C = 4.48 × 2 = 8.96litre

**Solution 29.**

**Solution 30.**

**Solution 31.**

**Solution 32.**

V_{1}/V_{2} = n_{1}/n_{2
}So, no. of moles of Cl = x/2 (since V is directly proportional to n)

No. of moles of NH_{3} = x

No. of moles of SO_{2} = x/4

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

So, 20 litre nitrogen contains x molecules

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

And 20 litre of ammonia will also contain =x molecules

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

**Solution 33.**

**Solution 34.**

(a) Volume of O_{2} = V

Since O_{2 }and N_{2 }have same no. of molecules = x

so, the volume of N_{2 }= V

(b) 3x molecules means 3V volume of CO

(c) 32 g oxygen is contained in = 44 g of CO_{2
}So, 8 g oxygen is contained in = 44 x 8/32 = 11 g

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

**Solution 35.**

(a) 444 g is the molecular formula of (NH_{4})_{2 }PtCl_{6
}% of Pt = (195/444) x 100 = 43.91% or 44%

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

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

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

So, the empirical formula is Na_{3}PO_{4}

**Solution 36.**

**Solution 37.
**

According to Avogadros law:

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

**Solution 38.**

**Solution 39.**

**Solution 40.**

**Solution 41.**

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

(ii) The volume will become double because volume is directly proportional to the no. of molecules at constant temperature and pressure.

V_{1}/V_{2} = n_{1}/n_{2
}V_{1}/V_{2} = n_{1}/2n_{1
}So, V_{2 }= 2V_{1}

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

(iv) The volume of D = 5.6 4 = 22.4 dm^{3}, so the number of molecules = 6 x 10^{23} because according to mole concept 22.4 litre volume at STP has = 6 x 10 ^{23} molecules

(v) No. of moles of D = 1 because volume is 22.4 litre

so, mass of N_{2}O = 1 44 = 44 g

**Solution 42.**

**Solution 43.**

**Solution 44.**

(a) Element % Atomic mass Atomic ratio Simple ratio

K 47.9 39 1.22 2

Be 5.5 9 0.6 1

F 46.6 19 2.45 4

so, empirical formula is K_{2}BeF_{4}

(b) 3CuO + 2NH_{3} → 3Cu + 3H_{2}O + N_{2
}3 V 2 V 3 V 1V

3 x 80 g of CuO reacts with = 2 x 22.4 litre of NH_{3
}so, 120 g of CuO will react with = 2x 22.4 x 120/80 x 3

= 22.4 litres

**Solution 45.**

(a) The molecular mass of ethylene(C_{2}H_{4}) is 28 g

No. of moles = 1.4/28 = 0.05 moles

No. of molecules = 6.023 x10^{23} x 0.05 = 3 x 10^{22 }molecules

Volume = 22.4 x 0.05 = 1.12 litres

(b) Molecular mass = 2 X V.D

S0, V.D = 28/2 = 14

**Solution 46.**

**Solution 47.**

**Solution 48.**

**Solution 49.**

**Solution 50.**

(a) (i) element % atomic mass at. ratio simple ratio

C 14.4 12 1.2 1

H 1.2 1 1.2 1

Cl 84.5 35.5 2.38 2

Empirical formula = CHCl_{2
}(ii) Empirical formula mass = 12+1+71= 84 g

Since molecular mass = 168 so, n = 2

so, molecular formula = (CHCl_{2})_{2} = C_{2}H_{2}Cl_{4}

(b) (i) C + 2H_{2}SO_{4} → CO_{2} + 2H_{2}O + 2SO_{2
}1 V 2 V 1 V 2 V

196 g of H_{2}SO_{4} is required to oxidized = 12 g C

So, 49 g will be required to oxidise = 49 x 12/196 = 3 g

(ii) 196 g of H_{2}SO_{4 }occupies volume = 2 x 22.4 litres

So, 49 g H_{2}SO_{4 }will occupy = 2 x 22.4 x 49/196 = 11.2 litre

i.e. volume of SO_{2} = 11.2 litre