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

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) N₂means 1 molecule of nitrogen and 2N means two atoms of nitrogen.
N₂ 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
nH₂=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.

2H₂ + O₂ → 2H₂O
2 V     1V         2V

From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm³ of Hydrogen reacts with = 200/2= 100 cm³
Hence, the unreacted oxygen is 150 – 100 = 50cm³ of oxygen.

Solution 5.

Solution 6.

Solution 7.

Solution 8.

Solution 9.

Solution 10.

Solution 11.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
1 V 5 V 3 V

From equation, 5 V of O₂ required = 1V of propane
so, 100 cm³ of O₂ will require = 20 cm³ of propane

Solution 12.

Solution 13.

2CO + O₂ → 2CO₂
2 V 1 V 2 V

2 V of CO requires = 1V of O₂
so, 100 litres of CO requires = 50 litre of O₂

Solution 14.

Solution 15.

H₂ + Cl₂ → 2HCl
1V 1V 2 V

Since 1 V hydrogen requires 1 V of oxygen and 4cm³ of H₂ remained behind so the mixture had com”>16 cm³ hydrogen and 16 cm³ chlorine.
Therefore Resulting mixture is H₂ =4cm³,HCl=32cm³

Solution 16.

CH₄ + 2O₂ → CO₂ + 2H₂O
1 V 2 V 1 V

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
2 V 5 V 4 V

From the equations, we can see that
1V CH₄ requires oxygen = 2 V O₂
So, 10cm³ CH₄ will require =20 cm³ O₂
Similarly 2 V C₂H₂ requires = 5 V O₂
So, 10 cm³ C₂H₂ will require = 25 cm³ O₂
Now, 20 V O₂ will be present in 100 V air and 25 V O₂ will be present in 125 V air ,so the volume of air required is 225cm³

Solution 17.

Solution 18.

Solution 19.

This experiment supports Gay lussac’s law of combining volumes.
Since the unchanged or remaining O₂ 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.

CH₄ + 2O₂ → CO₂ + 2H₂O
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 SO₂ 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 × 10²³
c) The molar volume of a gas at STP is 22.4 dm³ 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 dm³.

(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 x10²³ 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 :

1. It explains Gay-Lussac’s law.
2. It determines 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.

(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.

H₂ + Cl₂ → 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) KClO₃ = (K)39 + (Cl)35.5 + (3O)48 = 122.5
(c) (Cu)63.5 + (S)32 + (4O)64 + (5H₂O)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 x10²³ x 73/36.5(mol. mass of HCl)
= 12.04 x 10²³

(b) Weight of 0.5 mole of O₂ is = 32(mol. Mass of O₂) x 0.5=16 g

(c) No. of molecules in 1.8 g H₂O = 6.023 x 10²³ x 1.8/18
= 6.023 x 10²²

(d) No. of moles in 10g of CaCO₃ = 10/100(mol. Mass CaCO₃)
= 0.1 mole

(e) Weight of 0.2 mole H₂ gas = 2(Mol. Mass) x 0.2 = 0.4 g

(f) No. of molecules in 3.2 g of SO₂ = 6.023 x 10²³ x 3.2/64
= 3.023 x 10²²

Solution 6.

Molecular mass of H₂O is 18, CO₂ is 44, NH₃ is 17 and CO is 28
So, the weight of 1 mole of CO₂ is more than the other three.

Solution 7.

4g of NH₃ having minimum molecular mass contain maximum molecules.

Solution 8.

a) No. of particles in s1 mole = 6.023 x 10²³
So, particles in 0.1 mole = 6.023 x 10 ²³ x 0.1 = 6.023 x 10²²

b) 1 mole of H₂SO₄ contains =2 x 6.023 x 10²³
So, 0.1 mole of H₂SO₄ contains =2 x 6.023 x 10²³ x0.1
= 1.2×10²³ atoms of hydrogen

c) 111g CaCl₂ contains = 6.023 x 10²³ molecules
So, 1000 g contains = 5.42 x 10²⁴ 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 H₂O has mass = 0.2 x 18 = 3.6 g
(d) 0.1 mole of CO₂ 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 SO₂ has volume = 22.4 litres
So, 2 moles will have = 22.4 x 2 = 44.8 litre

Solution 11.

(a) 1 mole of CO₂ contains O₂ = 32g
So, CO₂ having 8 gm of O₂ 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 10²³ = 2.656 x 10^-23 g
(b) 1 atom of Hydrogen has mass = 1/6.023 x 10²³ = 1.666 x 10^-24
(c) 1 molecule of NH₃ has mass = 17/6.023 x10²³ = 2.82 x 10^-23 g
(d) 1 atom of silver has mass = 108/6.023 x 10²³ =1.701 x 10^-22
(e) 1 molecule of O₂ has mass = 32/6.023 x 10²³ = 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 CaCO₃ has mass =100(molar mass) x 0.1=10 g
(b) 0.1 mole of Na₂SO₄.10H₂O has mass = 322 x 0.1 = 32.2 g
(c) 0.1 mole of CaCl₂ 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 Na₂CO₃.10H₂O contains oxygen atoms = 13
So, 6.023 x10²³ molecules (1mole) has atoms=13 x 6.023 x 10²³
So, 0.1 mole will have atoms = 0.1 x 13 x 6.023 x 10²³ =7.8 x 10²³

Solution 15.

3.2 g of S has number of atoms = 6.023 x10²³ x 3.2 /32
= 0.6023 x 10²³
So, 0.6023 x 10²³ atoms of Ca has mass=40 x0.6023×10²³/6.023 x 10²³
= 4g

Solution 16.

(a) No. of atoms = 52 x 6.023 x10²³ = 3.131 x 10²⁵
(b) 4 amu = 1 atom of He
so, 52 amu = 13 atoms of He
(c) 4 g of He has atoms = 6.023 x10²³
So, 52 g will have = 6.023 x 10²³ x 52/4 = 7.828 x10²⁴ atoms

Solution 17.

Molecular mass of Na₂CO₃ = 106 g
106 g has 2 x 6.023 x10²³ atoms of Na
So, 5.3g will have = 2 x 6.023 x10²³x 5.3/106=6.022 x10²² atoms
Number of atoms of C = 6.023 x10²³ x 5.3/106 = 3.01 x 10²² atoms
And atoms of O = 3 x 6.023 x 10²³ x 5.3/106= 9.03 x10²² atoms

Solution 18.

(a) 60 g urea has mass of nitrogen(N₂) = 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 cm³ of CO has mass = 28 g
So, 56 cm³ will have mass = 56 x 28/22400 = 0.07 g

Solution 21.

18 g of water has number of molecules = 6.023 x 10²³
So, 0.09 g of water will have no. of molecules = 6.023 x 10²³ x 0.09/18 = 3.01 x 10²¹ molecules

Solution 22.

(a) No. of moles in 256 g S₈ = 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 10²³ = 1.2 x 10²² molecules
No. of atoms in 1 molecule of S = 8
So, no. of atoms in 1.2 x 10²² molecules = 1.2 x 10²² x 8
= 9.635x 10²² molecules

Solution 23.

Atomic mass of phosphorus P = 30.97 g
Hence, molar mass of P₄ = 123.88 g
If phosphorus is considered as P₄ molecules,
then 1 mole P₄ ≡ 123.88 g
Therefore, 100 g of P₄ = 0.807 g

Solution 24.

Solution 25.

No. of atoms in 12 g C = 6.023 x10²³
So, no. of carbon atoms in 10^-12 g = 10^-12 x 6.023 x10²³/12
= 5.019 x 10¹⁰ atoms

Solution 26.

Given:
P= 1140 mm Hg
Density = D = 2.4 g / L
T = 273 ⁰C = 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.

V₁= 1 L  T₁ = 546 K
V₂=?       T₂ = 273 K
V₁/T₁ = V₂/ T₂
V₂ = (V₁ x T₂)/T₁
      = (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  V₁ = 0.5 L
P₂ = 760 mm Hg  V₂ = ?
P₁ x V₁ = P₂ x V₂
V₂ = (P₁ x V₁)/P₂
      = (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 10²³ molecules
So, 4 g of NaOH contains = 6.02 x10²³ x 4/40
= 6.02 x10²² molecules

Solution 30.

The number of molecules in 18 g of ammonia= 6.02 x10²³
So, no. of molecules in 4.25 g of ammonia = 6.02 x 10²³ x 4.25/18
= 1.5 x 10²³

Solution 31.

(a) One mole of chlorine contains 6.023 x 10²³ 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 H₂O

1. That H₂O contains 2 volumes of hydrogen and 1 volume of oxygen.
2. That ratio by weight of hydrogen and oxygen is 1:8.
3. That molecular weight of H₂O 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) CH₂O (c) CH (d) CH₂O

Solution 4.

Solution 5.

Solution 6.

Molecular mass of KClO₃ = 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 Mg₃ N₂

Solution 18.

Barium chloride = BaCl₂.x H₂O
Ba + 2Cl + x[H₂ + O]
= 137+ 235.5 + x [2+16]
= [208 + 18x] contains water = 14.8% water in BaCl₂.x H₂O
= [208 + 18 x] 14.8/100 = 18x
= [104 + 9x] 2148=18000x
= [104+9x] 37=250x
= 3848 + 333x =2250x
1917x =3848
x = 2molecules of water

Solution 19.

Molar mass of urea; CON₂H₄ = 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 CH₂O
Since the compound has 12 atoms of carbon, so the formula is
C₁₂ H₂₄ 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 A₂B₄.

(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 A₆B₆

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 NH₂

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 =ZnSH₁₄O₁₁
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 = ZnSO₁₁H₁₄
= ZnSO₄.7H₂O

Exercise 5(D)

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Solution 5.

Molecular mass of KNO₃ = 101 g
63 g of HNO₃ is formed by = 101 g of KNO₃
So, 126000 g of HNO₃ is formed by = 126000 x 101/63 = 202 kg
Similarly,126 g of HNO₃ is formed by 170 kg of NaNO₃
So, smaller mass of NaNO₃ 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 = XO₂

Solution 10.

Solution 11.

Solution 12.

Solution 13.

(a) Number of molecules in 100cm³ 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 cm³ of nitrogen under the same conditions of temperature and pressure = Y
So, number of molecules in 50 cm³ 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 CH₃
(iii) The empirical formula mass for CH₂O = 30
V.D = 30
Molecular formula mass = V.D 2 = 60
Hence, n =mol. Formula mass/empirical formula mass= 2
So, molecular formula = (CH₂O)₂ = C₂H₄O₂

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 CO₂ = 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) V₁/V₂ = T₁/T₂
22.4/V₂ =273/546
V₂ = 44.8 litres
(d) Mass of 1 mole Cl₂ gas =35.5 x 2 =71 g

Solution 20.

Solution 21.

Solution 22.

(a) The molecular mass of (Mg(NO₃)₂.6H₂O = 256.4 g
% of Oxygen = 12 x 16/256
= 75%

(b) The molecular mass of boron in Na₂B₄O₇.10H₂O = 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 Cr₂O₃ = 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 273⁰C = 4.48 × 2 = 8.96litre

Solution 29.

Solution 30.

Solution 31.

Solution 32.

V₁/V₂ = n₁/n₂
So, no. of moles of Cl = x/2 (since V is directly proportional to n)
No. of moles of NH₃ = x
No. of moles of SO₂ = 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 O₂ = V
Since O₂ and N₂ have same no. of molecules = x
so, the volume of N₂ = V
(b) 3x molecules means 3V volume of CO
(c) 32 g oxygen is contained in = 44 g of CO₂
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 (NH₄)₂ PtCl₆
% 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 Na₃PO₄

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.
V₁/V₂ = n₁/n₂
V₁/V₂ = n₁/2n₁
So, V₂ = 2V₁

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

(iv) The volume of D = 5.6 4 = 22.4 dm³, so the number of molecules = 6 x 10²³ 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 N₂O = 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 K₂BeF₄

(b) 3CuO + 2NH₃ → 3Cu + 3H₂O + N₂
3 V 2 V 3 V 1V
3 x 80 g of CuO reacts with = 2 x 22.4 litre of NH₃
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(C₂H₄) is 28 g
No. of moles = 1.4/28 = 0.05 moles
No. of molecules = 6.023 x10²³ x 0.05 = 3 x 10²² 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 = CHCl₂
(ii) Empirical formula mass = 12+1+71= 84 g
Since molecular mass = 168 so, n = 2
so, molecular formula = (CHCl₂)₂ = C₂H₂Cl₄

(b) (i) C + 2H₂SO₄ → CO₂ + 2H₂O + 2SO₂
1 V 2 V 1 V 2 V
196 g of H₂SO₄ 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 H₂SO₄ occupies volume = 2 x 22.4 litres
So, 49 g H₂SO₄ will occupy = 2 x 22.4 x 49/196 = 11.2 litre
i.e. volume of SO₂ = 11.2 litre