In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking Kf for water as 1.85, the freezing point of the solution will be nearest to

Question

In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking Kf for water as 1.85, the freezing point of the solution will be nearest to (A) -0.480°C (B) -0.360°C (C) -0.260°C (D) + 0.480°C

Tardigrade Admin · Accepted Answer

HX leftharpoons H ++ X - i=(1-0.3+0.3+0.3/1)=1.3 Δ Tf=i Kf m =1.3 × 1.85 × 0.2=0.48 ∴ Tf=(0-0.48)° C =0-48° C.

Q. In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking $K_{f}$ for water as 1.85, the freezing point of the solution will be nearest to

-0.480 $^{\circ}$ C

-0.360 $^{\circ}$ C

-0.260 $^{\circ}$ C

+ 0.480 $^{\circ}$ C

$H X ⇌ H^{+} + X^{-}$
$i = \frac{1 - 0.3 + 0.3 + 0.3}{1} = 1.3$
$Δ T_{f} = i K_{f} m$
$= 1.3 \times 1.85 \times 0.2 = 0.48$
$∴ T_{f} = (0 - 0.48)^{\circ} C = 0 - 4 8^{\circ} C$ .

Q. In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3. Taking Kf​ for water as 1.85, the freezing point of the solution will be nearest to

-0.480∘C

-0.360∘C

-0.260∘C

+ 0.480∘C