Q8 / 90 · JEE Main 2012 (26 May Online) · Chemistry

The freezing point of a $1.00 \mathrm{~m}$ aqueous solution of $\mathrm{HF}$ is found to be $-1.91^{\circ} \mathrm{C}$. The freezing point constant of water, $K_f$ is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. The percentage dissociation of $\mathrm{HF}$ at this concentration is

  • A
    $30 \%$
  • B
    $10 \%$
  • C
    $5.2 \%$
  • Correct answer
    D
    $2.7 \%$

Solution

$\Delta T_{\mathrm{f}}=K_{\mathrm{f}} \times m \times i$ $$ \begin{aligned} & i=\frac{\Delta T_{\mathrm{f}}}{K_{\mathrm{f}} \times m}=\frac{1.91}{1.86 \times 1}=1.02 \\ & \text { For } \mathrm{HI} \leftrightharpoons \mathrm{H}^{+}+\mathrm{I}^{-} \\ & \quad(1-\alpha) \quad \alpha \quad \alpha \\ & 1-\alpha+\alpha+\alpha=i=1.027 \\ & 1+\alpha=1.027 \\ & \alpha=0.027 \text { or } 2.7 \% \end{aligned} $$

Hence, the correct answer is option: 'D'.

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