QJEEM21Ch315a: JEE Main 2021 Chemistry - Class 12 JEEM21Ch315a Solution: 224 mL of SO 2(g) at 298 K and 1 atm is passed through 100 mL…

moles of SO₂ = ${{224} \over {22400}}$ = 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 ${1 \over 2}$ $\times$ 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 = ${{36} \over {18}}$ = 2 moles

Accoding to Relative Lowering of Vapour :

${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}} + i{n_{N{a_2}C{O_3}}}}}$

[ ${{n_{N{a_2}C{O_3}}}}$ << ${{n_{{H_2}O}}}$ ]

${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}}}}$

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

$\Rightarrow$ 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