1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Where does the number 1.05 in your work come from? $\endgroup$ – Brinn Belyea Jul 18 '14 at 23:40
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.