A 0.50 mol kg^-1 aqueous solution of a weak monobasic acid HA is 20% dissociated (HA H+ + A-). Take K_f=1.86 and K_b=0.52 K kg mol^-1, with Delta T=iKm. Which statements are correct?

The molality of undissociated HA at equilibrium is 0.40 mol kg^-1

The solution boils at about 100.31 degC

The molality of undissociated HA at equilibrium is 0.40 mol kg^-1

The solution boils at about 100.31 degC

The solution freezes at about -0.93 degC

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3so182

A $0.50 mol k g^{- 1}$ aqueous solution of a weak monobasic acid HA is $20%$ dissociated ( $HA H X^{+} + A X^{-}$ ). Take $K_{f} = 1.86$ and $K_{b} = 0.52 K kg mo l^{- 1}$ , with $Δ T = i K m$ . Which statements are correct?

i = 1 + α = 1.20

Δ T_{b} = 1.20 \times 0.52 \times 0.50 = 0.312 K

, so it boils at

100.3 1^{\circ} C

m (HA) = 0.50 (1 - 0.20) = 0.40

Δ T_{f} = 1.20 \times 1.86 \times 0.50 = 1.116 K

, so it freezes at

- 1.1 2^{\circ} C

;

- 0.9 3^{\circ} C

would be obtained by ignoring dissociation.