An aqueous solution contains $27 g$ glucose (M $= 180$ ), $9.0 g$ urea (M $= 60$ ) and $17.1 g$ sucrose (M $= 342$ ) in $250 g$ of water (all in $g mo l^{- 1}$ ). None of the solutes dissociates. With $K_{f} = 1.86 K kg mo l^{- 1}$ , find the freezing point of the solution in $^{\circ} C$ , to two decimal places.

Numerical answer type, answer in degC, tolerance 0.01.

Answer: -2.6 degC

Colligative properties add over all non-volatile particles. Total moles

= 0.150 + 0.150 + 0.050 = 0.350 mol

0.250 kg

, so

m = 1.40 mol k g^{- 1}

Δ T_{f} = 1.86 \times 1.40 = 2.604 K

, giving a freezing point of

- 2.6 0^{\circ} C