Determine the freezing point of a solution containing $0.625 g$ of glucose $(C_{6} H_{12} O_{6})$ dissolved in $102.8 g$ of water.
(Freezing point of water $= 273 K$ , $K_{f}$ for water $= 1.87 K$ $k g$ ${m o l}^{- 1}$ , atomic weight $C = 12$ , $H = 1$ , $O = 16$ )

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Solution

The molecular weight of glucose $C_{6} H_{12} O_{6} = 6 (12) + 12 (1) + 6 (16) = 180$ g/mol.
The number of moles of glucose $= \frac{0.625 g}{180 g / m o l} = 0.00347 m o l e s$
Mass of water $= 102.8 g = 0.1028 k g$
The molality of glucose $m = \frac{0.00347 m o l e s}{0.1028 k g} = 0.0338 m o l / k g$
The depression in the freezing point $Δ T_{f} = K_{f} m = 1.87 K k g / m o l \times 0.0338 m o l / k g = 0.0632 K$
Freezing point of water = $273 K$
The freezing point of the solution $= 273 + 0.0632 = 273.0632 K$

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