A $\pu{4.305 g}$ sample of a nonelectrolyte is dissolved in $\pu{105 g}$ of water. The solution freezes at $\pu{-1.23^\circ C}$. Calculate the molar mass of the solute. $k_f$ for water is $\pu{1.86 C/m}$.

My steps:

$$1.23 = 1.86 \times \frac{x}{1.05},$$

where $x = y / 4.305$.

$$0.66 = \frac{x}{1.05}$$ $$x = 0.69$$ $$0.69 = y / 4.305$$ $$y = \pu{2.989 g}$$

But I know this is not correct.

  • $\begingroup$ Where does the number 1.05 in your work come from? $\endgroup$ – Brinn Belyea Jul 18 '14 at 23:40

The freezing point depression is based on the molal concentration (moles of solute per kg of solvent).

$$\Delta T_f = -k_f \cdot m$$

You know the freezing point depression of the solution and the cryoscopic constant, so you can calculate the molality:

$$m = -\dfrac{\Delta T_f}{k_f}=-\dfrac{-1.23 ^\circ \text{C}}{1.86 \frac{\text{kg}\cdot ^\circ \text{C}}{\text{mol}}}=0.66 molal$$

Since molality is the number of moles of solute divided by the mass of the solvent in kg, you can calculate the moles of the solute:

$$n_{solute}=0.66 m \cdot 0.105 \text {kg}= ...$$

Divide your mass of solute by the moles of solute and you will get the molar mass (and not 2.989).

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