A 0.20 m aqueous solution of a weak acid HA is 10% ionised (HA H+ + A-, K_f = 1.86 K kg mol^-1). Which of the following statements is/are correct?

The freezing point depression is 0.41 K (two decimals).

Its freezing point equals that of a completely dissociated 0.11 m NaCl solution.

The freezing point depression is 0.41 K (two decimals).

The molality of undissociated HA is 0.020 mol kg^-1.

Its freezing point equals that of a completely dissociated 0.11 m NaCl solution.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3so108

A $0.20 m$ aqueous solution of a weak acid HA is $10%$ ionised ( $HA H X^{+} + A X^{-}$ , $K_{f} = 1.86 K kg mo l^{- 1}$ ). Which of the following statements is/are correct?

AThe freezing point depression is $0.41 K$ (two decimals).
BThe molality of undissociated HA is $0.020 mol k g^{- 1}$ .
CThe van't Hoff factor is $1.10$ .
DIts freezing point equals that of a completely dissociated $0.11 m$ $NaCl$ solution.

i = 1 + α = 1.10

, so

Δ T_{f} = 1.10 \times 1.86 \times 0.20 = 0.409 K

. Particle molality

= 1.10 \times 0.20 = 0.22

, which equals

2 \times 0.11

for fully dissociated

NaCl

. Undissociated HA is

0.20 \times 0.90 = 0.18 mol k g^{- 1}

, not

0.020