Freezing Point

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Physical Chemistry › Freezing Point

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When 20 grams of a solute is added to a liter of water, the freezing point becomes $-2.3^{\circ} C.$ What is the molar mass of the solute?

Assume that the solute does not make ions in solution.

$K_f=1.86\frac{^{\circ}C}{m}$

$16.2\frac{g}{mol}$

$24.7\frac{g}{mol}$

$30.6\frac{g}{mol}$

$28.4\frac{g}{mol}$

We can use the freezing point depression equation in order to determine the molar mass of the unknown solute.

$\Delta T_{f} = K_{f}mi$

$2.3 = (1.86\frac{^{\circ}C}{m})(\frac{\frac{20grams}{x}}{1 liter})(1)$

$x = 16.2\frac{g}{mol}$