A 0.1 molal aqueous solution of a weak acid HA is $30 \%$ ionised. If $K_f$ for water is $1.86^{\circ}…

A 0.1 molal aqueous solution of a weak acid HA is $30 \%$ ionised. If $K_f$ for water is $1.86^{\circ} \mathrm{C} / \mathrm{m}$, the freezing point of the solution will be
  1. $-0.18^{\circ} \mathrm{C}$
  2. $-0.54^{\circ} \mathrm{C}$
  3. $-0.36^{\circ} \mathrm{C}$
  4. $-0.24^{\circ} \mathrm{C}$

Solution

Freezing point depression $\left(\Delta T_f\right)=i K_f m$ $\begin{aligned} & i=1-0.3+0.3+0.3 \\ & i=1.3 \\ & \therefore \quad \Delta T_f=1.3 \times 1.86 \times 0.1=0.2418^{\circ} \mathrm{C} \\ & T_f=0-0.2418^{\circ} \mathrm{C} \\ & =-0.2418^{\circ} \mathrm{C} \\ & \end{aligned}$

Asked in: NEET 2011 (Mains)

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