A $1.00 mol L^{- 1}$ aqueous glucose solution (molar mass $180 g mo l^{- 1}$ ) has a density of $1.12 g m L^{- 1}$ . Using $K_{f} = 1.86 K kg mo l^{- 1}$ , calculate the depression in freezing point in kelvin, to two decimal places.

Numerical answer type, answer in K, tolerance 0.01.

Answer: 1.98 K

Take

1 L

: mass of solution

= 1120 g

, glucose

= 180 g

, so water

= 940 g

. Molality

= 1.00/0.940 = 1.0638 mol k g^{- 1}

. Then

Δ T_{f} = 1.86 \times 1.0638 = 1.9787 \approx 1.98 K