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

Tardigrade Admin · Accepted Answer

Freezing point depression (Δ Tf)=i Kf m <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/eb70428298bd593bbcac038ac9e1ae36-.png" /> ∴ Δ Tf =1.3 × 1.86 × 0.1=0.2418° C Tf =0-0.2418° C =-0.2418° C

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

$- 0.1 8^{\circ} C$

$- 0.5 4^{\circ} C$

$- 0.3 6^{\circ} C$

$- 0.2 4^{\circ} C$

Freezing point depression $(Δ T_{f}) = i K_{f} m$

$∴ Δ T_{f} = 1.3 \times 1.86 \times 0.1 = 0.241 8^{\circ} C$
$T_{f} = 0 - 0.241 8^{\circ} C$
$= - 0.241 8^{\circ} C$

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

−0.18∘C

−0.54∘C

−0.36∘C

−0.24∘C