An aqueous solution contains $0.10 mol k g^{- 1}$ $NaCl$ (completely dissociated) and $0.20 mol k g^{- 1}$ glucose. Use $K_{f} = 1.86$ and $K_{b} = 0.52 K kg mo l^{- 1}$ . Which statements are correct?

AThe solution freezes at $- 0.7 4^{\circ} C$
BThe solution boils at $100.4 2^{\circ} C$
CThe freezing point is $- 0.5 6^{\circ} C$
DIf $NaCl$ is replaced by the same molality of completely dissociated $CaCl X_{2}$ , the depression of freezing point becomes $0.93 K$

Answer: A, D

Particle molality

= 2 (0.10) + 0.20 = 0.40

Δ T_{f} = 1.86 \times 0.40 = 0.744 K

so the freezing point is

- 0.7 4^{\circ} C

, and

Δ T_{b} = 0.52 \times 0.40 = 0.208 K

so the boiling point is

100.2 1^{\circ} C

. With

CaCl X_{2}

the particle molality is

0.30 + 0.20 = 0.50

, so

Δ T_{f} = 0.93 K

. The values

100.42

and

- 0.56

come from doubling

Δ T_{b}

and from ignoring the ionisation of

NaCl