0.5 molal aqueous solution of a weak acid (HX) is 20\% ionized. If K_f for water is 1.86\ K kg mol^ -1 Calculate the lowering in freezing point of the solution.

Correct answer: B. 1.12\ K The ionization of the weak acid HX is given by: HX leftharpoons H^+ + X^- The van't Hoff factor i is calculated as: i = 1 + α Given that the acid is 20\% ionized, α = 0.2 . i = 1 + 0.2 = 1.2 The lowering in freezing point Δ T_f is given by: Δ T_f = i × K_f × m Substituting the given values: Δ T_f = 1.2 × 1.86 × 0.5 Δ T_f = 1.116\ K 1.12\ K Answer: 1.12\ K

MHT CET 202621 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 21 April
Shift: Evening Shift
Subject: Chemistry
Chapter: Solutions

MHT CET 2026 · 21 April · Evening Shift · Chemistry · Solutions

0.5 molal aqueous solution of a weak acid (HX) is 20\% ionized. If...

0.5

molal aqueous solution of a weak acid (HX) is

20\%

ionized. If

K_f

for water is

1.86\ \text{K kg mol}^{-1}

Calculate the lowering in freezing point of the solution.

Options

A $0.56\ \text{K}$
B $1.12\ \text{K}$
C $-0.56\ \text{K}$
D $-1.12\ \text{K}$

Correct answer

B. $1.12\ \text{K}$

Step-by-step solution

The ionization of the weak acid

\text{HX}

is given by:

\text{HX} \rightleftharpoons \text{H}^+ + \text{X}^-

The van't Hoff factor

i

is calculated as:

i = 1 + \alpha

Given that the acid is

20\%

ionized,

\alpha = 0.2

i = 1 + 0.2 = 1.2

The lowering in freezing point

\Delta T_f

is given by:

\Delta T_f = i \times K_f \times m

Substituting the given values:

\Delta T_f = 1.2 \times 1.86 \times 0.5

\Delta T_f = 1.116\ \text{K} \approx 1.12\ \text{K}

Answer:

1.12\ \text{K}

Practice Solutions on Quantrex Academy →

More from Solutions