MCQ

For each of the options, we will calculate the number of moles.

At STP, 22.4 L = 1 mol

Moles of H₂(g) =  = 0.67 mol

moles of N₂ (g)=  = 0.223 mol
Moles of 0.5 g H₂ gas =  = 0.25 mol

moles of 10 g O₂ gas =  = 0.1325 mol

Maximum number of moles = maximum number of molecules = option (a)

Maximum number of moles have maximum number of molecules.
Moles of N₂O = = 0.16 mol 

Moles of H₂ = = 10 mol

Moles of NO₂ = = 0.35 mol
Moles of SO₂ = = 0.25 mol

H₂ has maximum number moles and thus maximum number of molecules. Correct option is (b).

Avogadro number = N_o

_{Total number of e}^_ _{in CH4 = 10}

Number of e^– in 1.6 g of CH₄ = = N_o

_{Total number of e}^_ _{in H2O = 10}

Number of e^– in 1.8 g of H₂O = = N_o

So, correct option is (b).

Maximum number of moles have maximum number of molecules
So, we will calculate number of moles for each of these.
8 g H₂ =  = 4 moles 

44 g of CO₂ =  = 1 mole
64 g SO₂ =  = 1 mole

48 g of O₃ =  = 1 mole

Option (a) is correct

molar mass of CuSO₄.5H₂O = 249.5 g/mol
1 mole = 6.023 x 10²³ molecules
Mass of 6.023 x 10²³ molecules = 249.5 g
mass of 1 x 10²² molecules =
Correct option is (b).

The best way is to convert all the options into the moles.
Option (a) 10.8 g of Ag: 108 g of Ag = 1 mole
10.8 g of Ag = 10.8/108 = 0.1 mol
Option (b) 2.4 of C: 12 g of C = 1 mole
2.4 g of C = 2.4/12 = 0.2 mol
Option (c) 5.6 g of Fe: 56 g of Fe = 1 mole
5.6 g of Fe = 5.6/56 = 0.1 mol
Option (d) 54 g of Al: 27 g of Al = 1 mole
54 g of Al = 54/27 = 2.0 mol
More is the number of moles, more are the number of atoms.
So, correct option is (d).

1 mole of methane = 16 g = 6.02 x 10²³ molecules of CH₄
1.6 g = (1.6/16) x 6.02 x 10²³ = 6.02 x 10²² molecules of CH₄
C has 6 electrons and H has 1 electron.
In total, 1 molecule of CH₄ has 10 electrons.
So, 6.02 x 10²² molecules of CH₄ will have 6.02 x 10²² x 10 electrons = 6.02 x 10²³ electrons
Correct option is (b).

For option (a),
1 mole of CO₂ = 44 g
then, 0.2 mol of CO₂ = 0.2 x 44 = 8.8 g
For option (b),
22.4 L of SO₂ = 1 mole = 64 g
2.24 L of SO₂ = (2.24/22.4) x 64 = 6.4 g
For option (c),
6.02 x 10²³ molecules of water = 1 mole = 18 g
3.01 x 10²³ molecules of water = (3.01/6.02) x 18 = 9 g
For option (d),
mass given = 20 g calcium
So, correct option is (d).

The equation for reaction of coal with oxygen during burning is

1 mole of C reacts with 1 mole of O₂.
Mass of coal = 1 kg = 1000 g
Molar mass of C = 12 g/mol
Moles of C = mass/molar mass = 1000/12 = 83.3 mol
83.3 moles of carbon will react with 83.3 moles of oxygen.
At NTP, 1 mol = 22.4 L
83.3 moles = 22.4 x 83.3 = 1866 L = 1.86 x 10³ L
Option (b) is correct.

Mass of KI = 3.69 g
Mass of water = 21.70 g
Total mass of solution = 3.69+21.70 = 25.39 g
Density of solution = 1.11 g/ml
Total volume of solution = mass of solution/density = 25.39/1.11 = 22.87 ml
So, 22.87 ml of solution contains 3.69 g of KI
100 ml of solution will contain mass of KI = 3.69 x 100/22.87 = 16.13 g
So, % w/v of KI is 16.1 % (w/v)
Option (b) is correct.