A solution of $36 g$ glucose (M $= 180 g mo l^{- 1}$ ) in $400 g$ of water is slowly cooled to $- 2.7 9^{\circ} C$ , and pure ice separates until the remaining liquid solution is in equilibrium at that temperature. Taking $K_{f} = 1.86 K kg mo l^{- 1}$ and ideal behaviour, find the mass of ice that has separated, in grams, to the nearest integer.

Numerical answer type, answer in g, tolerance 1.

Answer: 267 g

- 2.7 9^{\circ} C

Δ T_{f} = 2.79 K

, so the equilibrium molality is

m^{'} = 2.79/1.86 = 1.50 mol k g^{- 1}

. All

0.200 mol

of glucose stays in the liquid, so liquid water

= 0.200/1.50 = 133.33 g

(after multiplying by 1000). Ice

= 400 - 133.33 = 266.67 \approx 267 g