# Question 5e092

Sep 20, 2015

The molar mass of the compound is $\text{330 g/mol}$.

#### Explanation:

The way I see it, the only reasonable thing to determine here is the molar mass of the unknown compound.

The idea is that you can use the freezing point of the solution to find its molality. The equation for freezing-point depression looks like this

$\Delta {T}_{f} = i \cdot {K}_{f} \cdot b \text{ }$, where

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor, equal to $1$ for non-electrolytes;
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

The cryoscopic constant of water is equal to $1.853 {\text{^@"C kg mol}}^{- 1}$. The freezing-point depression is defined as

$\Delta {T}_{f} = {T}_{f}^{\circ} - {T}_{f} \text{ }$, where

${T}_{f}$ - the freezing point of the solution;
${T}_{f}^{\circ}$ - the freezing point of the pure solvent.

$\Delta {T}_{f} = 0 \text{^@"C" - (-1.3""^@"C") = 1.3""^@"C}$

This means that the molality of the solution will be

$b = \frac{\Delta {T}_{f}}{i \cdot {K}_{f}}$

b = (1.3color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.853color(red)(cancel(color(black)(""^@"C"))) "Kg mol"""^(-1)) = "0.7016 molal"#

Molality is defined as moles of solute, in your case the unknown non-electrolyte, divided by the mass of the solvent - expressed in kilograms.

Use the solution's molality to find the number of moles of solute

$b = {n}_{\text{solute"/m_"solvent" implies n_"solute" = b * m_"solvent}}$

${n}_{\text{solute" = 0.7016"mol"/color(red)(cancel(color(black)("kg"))) * 156.5 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.1098 moles}}$

This means that the molar mass of the compound will be

${M}_{M} = \frac{m}{n} = \text{35.9 g"/"0.1098 moles" = "326.9 g/mol}$

Rounded to two sig figs, the answer will be

${M}_{M} = \textcolor{g r e e n}{\text{330 g/mol}}$

SIDE NOTE If this is not what you are looking for, let me know, I will wdit the answer accordingly.