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Q.

The freezing point of a 1.00m aqueous solution of HF is found to be $− 1.91 ∘ C$ . The freezing point constant of water, $K_{f} is 1.86 {K.kg.mol}^{- 1}$ . The percentage dissociation of HF at this concentration is

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a

5.2%

b

30%

c

10%

d

2.7%

answer is D.

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Detailed Solution

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$Δ T_{f} = i \times K_{f} \times m, i = \frac{Δ T_{f}}{K_{f} \times m} = \frac{1.91}{1.86 \times 1} = 1.027 For H F ⇌ H^{+} + F^{-}$
$1 - α + α + α = i = 1.027 \Rightarrow α = 0.027 or 2.7 %$

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