A 0.1 molal aqueous solution of a weak acid is 30 \% ionised. If K_f for water is 1.86^ C / m , the freezing point of the solution will be

Correct answer: D. -0.24^ C Freezing point depression (Δ T_f)=i K_f m aligned & i=1-0.3+0.3+0.3 & i=1.3 & Δ T_f=1.3 ×1.86 ×0.1=0.2418^ ^ C & T_f=0-0.2418^ ^ C & =-0.2418^ ^ C & aligned

NEET 2011

Exam: NEET
Year: 2011
Subject: Chemistry
Chapter: Solutions

NEET 2011 · Chemistry · Solutions

A 0.1 molal aqueous solution of a weak acid is 30 \% ionised. If...

A 0.1 molal aqueous solution of a weak acid is

30 \%

ionised. If

K_f

for water is

1.86^{\circ} \mathrm{C} / \mathrm{m}

, the freezing point of the solution will be

Options

A $-0.18^{\circ} \mathrm{C}$
B $-0.54^{\circ} \mathrm{C}$
C $-0.36^{\circ} \mathrm{C}$
D $-0.24^{\circ} \mathrm{C}$

Correct answer

D. $-0.24^{\circ} \mathrm{C}$

Step-by-step 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 \times1.86 \times0.1=0.2418^{{}^{\circ}} \mathrm{C}\\ & T_f=0-0.2418^{{}^{\circ}} \mathrm{C}\\ & =-0.2418^{{}^{\circ}} \mathrm{C}\\ & \end{aligned}

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