A solution contains $0.30 mol$ of a non-volatile, non-dissociating solute in $500 g$ of water. Use $K_{f} = 1.86 K kg mo l^{- 1}$ and $K_{b} = 0.52 K kg mo l^{- 1}$ and treat the solution as ideal at all concentrations. Which statements are correct?

Answer: A, B, C

m = 0.30/0.500 = 0.60

Δ T_{f} = 1.86 \times 0.60 = 1.116

so the freezing point is

- 1.1 2^{\circ} C

. At

Δ T_{f} = 2.232 K

the equilibrium molality is

2.232/1.86 = 1.20

, so liquid water

= 0.30/1.20 = 0.250 kg

, which is half of the original

0.500 kg

. For boiling,

Δ T_{b} = 0.52 \times 0.60 = 0.312

, giving

100.3 1^{\circ} C

;

100.62

would result from doubling.