moles of SO₂ = = 0.01

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

The balanced equation is

SO₂ + 2NaOH Na₂SO₃ + H₂O

Here NaOH is limiting Reagent.

2 mole NaOH produces 1 mole Na₂SO₃

0.01 mole NaOH produces 0.01 mole Na₂SO₃

Moles of Na₂SO₃ = 0.005 mole

Na₂SO₃ 2Na⁺ + SO₃^2-

van’t Hoff factor (i) = 3

Moles of H₂O = = 2 moles

Accoding to Relative Lowering of Vapour :



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24 – P_S = 0.18

P_S = 23.82

Lowering in pressure (P) = 0.18 mm of Hg

= 18 × 10^–2 mm of Hg