1.5 g of a monoprotic acid when dissolved in 150 g of water lowers the freezing point by 0.9 3 ∘ C .0.5 g of the same acid when titrated, after dissolution in water, requires 125 ml of N /10 alkali. C

Question

1.5 g of a monoprotic acid when dissolved in 150 g of water lowers the freezing point by 0.9 3 ∘ C .0.5 g of the same acid when titrated, after dissolution in water, requires 125 ml of N /10 alkali. Calculate the percentage dissociation of acid in this solution ( K f ​ for water = 1.86 K kg mol − 1 ) .

TestHub · Accepted Answer

From titration, moles of acid = moles of alkali.

$(0.5 g / M) = (1/10) N \times 0.125 L ⟹ M = 40 g / m o l$

From freezing point depression, $Δ T_{f} = i K_{f} m$ .

$0.93 = i \times 1.86 \times (1.5 g /40 g / m o l) / (0.150 k g) ⟹ i = 2$

Since $i = 1 + α$ for a monoprotic acid, $2 = 1 + α ⟹ α = 1$ .

Percentage dissociation is $100%$ .

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