A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86° C/m, the freezing point of the solution will be -

Question

A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86° C/m, the freezing point of the solution will be - (A) -0.24° C (B) -0.18° C (C) -0.54° C (D) -0.36° C

Tardigrade Admin · Accepted Answer

i = 1 - α + nα i = 1 - 0.3 + 2 (0.3) i = 1.3 Δ Tf = iKf m = 1.3 × 1.86 × 0.1 Δ Tf = + 0.24° C Freezing point of solution = - 0.24ºC

Q. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If $K_{f}$ for water is $1.8 6^{\circ} C / m$ , the freezing point of the solution will be -

$- 0.2 4^{\circ} C$

$- 0.1 8^{\circ} C$

$- 0.5 4^{\circ} C$

$- 0.3 6^{\circ} C$

$i = 1 - α + n α$
$i = 1 - 0.3 + 2 (0.3)$
$i = 1.3$
$Δ T_{f} = i K_{f} m$
$= 1.3 \times 1.86 \times 0.1$
$Δ T_{f} = + 0.2 4^{\circ} C$
Freezing point of solution = - 0.24ºC

Q. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf​ for water is 1.86∘C/m, the freezing point of the solution will be -

−0.24∘C

−0.18∘C

−0.54∘C

−0.36∘C