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A dissolution has $10$g of some unknown liquid and also $90$g of water. The freezing point of this dissolution is $-3.33^\text{o}$C. What is the molar mass of the unknown liquid in g/mol?

Alright. First of all, the formula to calculate the freezing point of a dissolution is

$$T_f - \Delta_f$$

Where

$$\Delta_f = \text{m} \cdot K_f$$

In our case, it would be

$$-3.33 = 0 - (\text{m}\cdot 1.86)$$

The molality m is

$$\frac{\text{moles of unknown liquid}}{\text{kg of water}} = \frac{\text{moles of unknown liquid}}{0.09}$$

We have to solve the equation:

$$-3.33=-\frac{\text{moles of unknown liquid}}{0.09}\cdot 1.86$$

Which yields

$$0.16 = \text{moles of unknown liquid}$$

However, I need the molar mass of this unknown liquid. How do I get it?

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    $\begingroup$ Well you know that you have 0.16 moles of the liquid. What else do you know about the liquid? $\endgroup$ – MaxW Nov 5 '15 at 22:29
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As long as you work with sufficently low concentrations, you do not need the molecular mass of your unkown. Indeed, this was (is?) your initial question.

Freezeing point depression (salting the footwalk to melt ice in the winter), elevation of the boiling point (adding salt to cooking water that increases it's boiling temperature, too) and osmotic pressure are colligative properties. What you need is to look up the crysoscopic constant, for water (same page) of $1.853 \mathrm{~K·kg/mol}$.

On the other side, if you already state the equation of $0.16$ moles corresponding to $10\mathrm{~g}$.

Despite advent of mass spectroscopy, such a determination may be still helpful today.

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  • $\begingroup$ So low that it doesn't really change the melting point? Because if it changes it at all, something that weighs 50 g/mol will have twice the effect of something that weighs 100 g/mol, assuming the same mass. $\endgroup$ – SendersReagent Mar 5 '16 at 1:44

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