Question

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

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

Answer 1

The Correct Option is A

$i = 1 - \alpha + n\alpha$
$i = 1 - 0.3 + 2 (0.3)$
$i = 1.3$
$\Delta T_f = iK_f m$
$= 1.3 \times 1.86 \times 0.1$
$\Delta T_f = + 0.24^{\circ} C$
Freezing point of solution = - 0.24�C