Mole Concept And Stoichiometry Exe-5(B) Concise Class-10 ICSE Chemistry Selina Solutions Chapter-5. We Provide Step by Step Answer of Exe-5(B) Questions of Exercise-5 for ICSE Class-10. The given Solutions is according to the Latest editions. Visit official Website CISCE for detail information about ICSE Board Class-10.

Mole Concept And Stoichiometry Exe-5(B) Concise Class-10 ICSE Chemistry Selina Solutions Chapter-5

Board	ICSE
Book / Publication	Concise / Selina
Subject	Chemistry
Class	10th
Writer	Dr SP Singh
Chapter-5	Mole Concept And Stoichiometry
Topics	Exe-5(B)
Edition	2025-2026

Exe-5(B) Questions on Mole Concept And Stoichiometry

Page-85

Que-1: Calculate the relative molecular masses of:

(a) Ammonium chloroplatinate, (NH₄)₂ PtCl₆
(b) Potassium chlorate
(c) CuSO₄. 5H₂O
(d) (NH₄)₂SO₄
(e) CH₃COONa
(f) CHCl₃
(g) (NH₄)₂ Cr₂O₇

Ans: 

(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

Que-2: Find the

(a) number of molecules in 73 g of HCl,
(b) weight of 0.5 mole of O₂,
(c) number of molecules in 1.8 g of H₂O
(d) number of moles in 10 g of CaCO₃
(e) Weight of 0.2 mole of H₂ gas,
(f) Number of molecules in 3.2 g of SO_2.

Ans: 

(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

Que-3: Which of the following would weigh most?

(a) 1 mole of H₂O
(b) 1 mole of CO₂
(c) 1 mole of NH₃
(d) 1 mole of CO

Ans: 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.

Que-4: Which of the following contains maximum number of molecules?

(a) 4 g of O₂
(b) 4 g of NH₃
(c) 4 g of CO₂
(d) 4 g of SO₂

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

Que-5: Calculate the number of

(a) Particles in 0.1 mole of any substance.
(b) Hydrogen atoms in 0.1 mole of H₂SO₄.
(c) Molecules in one Kg of calcium chloride.

Ans: 

(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

Que-6: How many grams of

(a) Al are present in 0.2 mole of it?
(b) HCl are present in 0.1 mole of it?
(c) H₂O are present in 0.2 mole of it?
(d) CO₂ is present in 0.1 mole of it?

Ans: 

(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

Que-7: 

(a) The mass of 5.6 litres of a certain gas at S.T.P. is 12 g. What is the relative molecular mass or molar mass of the gas? (2017)
(b) Calculate the volume occupied at S.T.P. by 2 moles of SO₂.

Ans: 

(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

Que-8: Calculate the number of moles of

(a) CO₂ which contain 8.00 g of O₂
(b) Methane in 0.80 g of methane.

Ans: 
(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

Que-9: Calculate the actual mass of

(a) An atom of oxygen
(b) an atom of hydrogen
(c) a molecule of NH₃
(d)  10²² atoms of carbon
(e) the molecule of oxygen
(f) 0.25 gram atom of calcium

Ans: 

(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) Mass of 6.022 × 10²³ atoms of atomic carbon = 12 g
∴ Mass of 10²² atoms of carbon = (12/6.022×10²³ )x10²²
= 0.2 g
(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

Que-10: Calculate the mass of 0.1 mole of each of the following

(a) CaCO₃
(b) Na₂SO₄.10H₂O
(c) CaCl₂
(d) Mg
(Ca = 40, Na=23, Mg =24, S=32, C = 12, Cl = 35.5, O=16, H=1)

Ans: 

(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

Que-11: Calculate the number of

(a) oxygen atoms in 0.10 mole of Na₂CO₃.10H₂O. (2017)
(b) gram atoms in 4.6 gram of sodium (2019)
(c) Mole in 12 g of oxygen

Ans: 

(a) 1 molecule 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 =7.8×10²³
(b) Given Na = 4.6 gm
At. mass = 23
No. of gram atoms of Na=    Mass of Na
At. mass of Na
= 4.6
23
= 0.2 gms
(c) 32 g of oxygen gas = 1 mole
1 gram of oxygen gas = 1/32 mole
Given that 12 g of oxygen gas
No: of moles = given mass / molar mass
= 12 / 32  = 0.375 mole

Que-12: What mass of Ca will contain the same number of atoms as are present in 3.2 g of S? (2015)

Ans: 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
x10²³
= 4g

Que-13: Calculate the number of atoms in each of the following:

(a) 52 moles of He
(b) 52 amu of He
(c) 52 g of He

Ans: 

(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

Que-14: Calculate the number of atoms of each kind in 5.3 grams of sodium carbonate.

Ans: 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

Que-15: 

(a) Calculate the mass of nitrogen supplied to soil by 5 kg of urea [CO(NH₂)₂] [O = 16; N = 14; C = 12 ; H = 1 ]
(b) Calculate the volume occupied by 320 g of sulphur dioxide at S.T.P. [S = 32; O = 16]

Ans: 

(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

Que-16: 

(a) What do you understand by the statement that ‘vapour density of carbon dioxide is 22’?
(b) Atomic mass of Chlorine is 35.5.What is its vapour density?

Ans: 

(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.

Que-17: What is the mass of 56 cm³ of carbon monoxide at STP?

(C=12 ,O=16)

Ans: 22400 cm³ of CO has mass = 28 g
So, 56 cm³ will have mass = 56 x 28/22400 = 0.07 g

Que-18: Determine the number of molecules in a drop of water which weighs 0.09g.

Ans: 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

Que-19: 

The molecular formula for elemental sulphur is S₈.In sample of 5.12 g of sulphur
(a) How many moles of sulphur are present?
(b) How many molecules and atoms are present?

Ans: (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

Que-20: If phosphorus is considered to contain P₄ molecules, then calculate the number of moles in 100g of phosphorus?

Ans: 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

Que-21:  Calculate:

(a) The gram molecular mass of chlorine if 308cm³ of it at STP weighs 0.979 g
(b) The volume of 4g of H₂ at 4 atmospheres.
(c) The mass of oxygen in 2.2 litres of CO₂ at STP.

Ans: (a) 308 cm³ of chlorine weighs = 0.979 g
So, 22400 cm³ will weigh = gram molecular mass
= 0.979 x 22400/308 =71.2 g
(b) 2 g(molar mass) H₂ at 1 atm has volume = 22.4 litres
So, 4 g H₂ at 1 atm will have volume = 44.8 litres
Now, at 1 atm(P₁) 4 g H₂ has volume (V₁) = 44.8 litres
So, at 4 atm(P₂) the volume(V₂) will be =
(c) Mass of oxygen in 22.4 litres = 32 g(molar mass)
So, mass of oxygen in 2.2 litres = 2.2 x 32/22.4=3.14 g

Que-22: A student puts his signature with graphite pencil. If the mass of carbon in the signature is 10^-12 g, calculate the number of carbon atoms in the signature.

Ans: 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

Que-23: An unknown gas shows a density of 3 g per litre at 273⁰C and 1140 mm Hg pressure. What is the gram molecular mass of this gas?

Ans: 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 litre 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

Que-24: Cost of Sugar (C₁₂H₂₂ O₁₁) is Rs 40 per kg; calculate its cost per mole.

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

Que-25: Calculate the number of molecules in one kg of NaOH.

Ans: Mass of 1 mole of NaOH = Na + O + H = 23 + 16 + 1 = 40 g
40 g of NaOH contains 6.022 × 10²³ molecules
∴ 1000 g of NaOH contains = (6.022×10²³ ×1000)/40​
= 1.5 × 10²⁵ molecules

Que-26: Calculate the number of atoms present in :

(a) 10 g of Chlorine
(b) 10 g of Nitrogen

Ans:
(a) Mass of 1 mole of chlorine is 35.5 g
35.5 g of chlorine contains 6.022 × 10²³ atoms
∴ 10g of chlorine contains = (6.022×10²³×10)/35.5​
= 1.7 x 10²³ atoms
(b) Mass of 1 mole of nitrogen is 14 g
14 g of nitrogen contains 6.022 × 10²³ atoms
∴ 10g of nitrogen contains = (6.022×10²³×10)/14​
= 4.3 x 10²³ atoms

Que-27: Correct the following:

(a) Equal volumes of any gas, under similar conditions, contain an equal number of atoms.
(b) 22 g of CO₂, occupies 22.4 litres at STP.
(c) The unit of atomic weight is grams.

Ans:
(a) Equal volumes of any gas, under similar conditions, contain an equal number of molecules.
(b) 44 g of CO₂, occupies 22.4 litres at STP.
(c) The unit of atomic weight is atomic mass unit (a.m.u).

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