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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
$-0.36^\circ C$
B
$-0.24^\circ C$
C
$-0.18^\circ C$
D
$-0.54^\circ C$
Detailed Solution
A weak monobasic acid ionises as $HA \rightleftharpoons H^+ + A^-$, giving n = 2 particles per molecule.
Degree of ionisation $\alpha = 0.30$.
van't Hoff factor: $i = 1 + (n - 1)\alpha = 1 + (2 - 1)(0.30) = 1.3$
$\Delta T_f = i\times K_f\times m = 1.3\times1.86\times0.1$
$\Delta T_f = 0.2418^\circ C$
Freezing point of the solution = $0 - 0.24 = -0.24^\circ C$
Degree of ionisation $\alpha = 0.30$.
van't Hoff factor: $i = 1 + (n - 1)\alpha = 1 + (2 - 1)(0.30) = 1.3$
$\Delta T_f = i\times K_f\times m = 1.3\times1.86\times0.1$
$\Delta T_f = 0.2418^\circ C$
Freezing point of the solution = $0 - 0.24 = -0.24^\circ C$