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?