JEE Main 2021 — Solutions Question with Solution

moles of SO₂ = $$\frac{224}{22400}$$ = 0.01

moles of NaOH = molarity × volume (in litre)
= 0.1 × 0.1
= 0.01 moles

The balanced equation is

SO₂ + 2NaOH $$\to$$ Na₂SO₃ + H₂O

$$\therefore$$ Here NaOH is limiting Reagent.

2 mole NaOH produces 1 mole Na₂SO₃

0.01 mole NaOH produces $$\frac{1}{2}$$ $$\times$$ 0.01 mole Na₂SO₃

$$\therefore$$ Moles of Na₂SO₃ = 0.005 mole

Na₂SO₃ $$\to$$ 2Na⁺ + SO₃^2-

van’t Hoff factor (i) = 3

Moles of H₂O = $$\frac{36}{18}$$ = 2 moles

Accoding to Relative Lowering of Vapour :

$$\frac{P_{H_{2}O}^o-P_S}{P_{H_{2}O}^o}=\frac{i{n}_{N{a}_{2}C{O}_3}}{n_{H_{2}O}+i{n}_{N{a}_{2}C{O}_3}}$$

[ $$n_{N{a}_{2}C{O}_3}$$ << $$n_{H_{2}O}$$ ]

$$\frac{P_{H_{2}O}^o-P_S}{P_{H_{2}O}^o}=\frac{i{n}_{N{a}_{2}C{O}_3}}{n_{H_{2}O}}$$

$$\Rightarrow$$ $$\frac{24-P_S}{24}=\frac{3\times 0.005}{2}$$

$$\Rightarrow$$ 24 – P_S = 0.18

$$\Rightarrow$$ P_S = 23.82

Lowering in pressure ($$\Delta$$P) = 0.18 mm of Hg

= 18 × 10^–2 mm of Hg