# Question ec1e7

Sep 17, 2015

$\text{159 g/mol}$

#### Explanation:

The idea here is that you need to use the difference between the freezing point of this solution and the freezing point of pure cyclohexane to determine the molality of the solution.

This, in turn, will help you find the number of moles of solute.

So, you know that the equation that links freezing-point depression and molality looks like this

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

$\Delta {T}_{f}$ - the freezing-pointdepression;
$i$ - 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 freezing point of pure cyclohexane is ${6.47}^{\circ} \text{C}$ and its cryoscopic constant is equal to ${\text{20.2 kg" ""^@"C mol}}^{- 1}$.

Rearrange the equation and solve for $b$ to get

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

In this case, the freezing-point depression will be

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

$\Delta {T}_{f} = 6.47 \text{^@"C" - 2.79""^@"C" = 3.98""^@"C}$

This means that $b$ is equal to

b = (3.98color(red)(cancel(color(black)(""^@"C"))))/(1 * "20.2 kg"color(red)(cancel(color(black)(""^@"C")))"mol"""^(-1)) = "0.1970 mol/kg"

Since molality is defined as moles of solute per kilograms of solvent, you can say that

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

This means that you have

${n}_{\text{solute" = 0.1970"mol"/color(red)(cancel(color(black)("kg"))) * 20.0 * 10^(-3)color(red)(cancel(color(black)("kg"))) = "0.00394 moles}}$

Use the compound's known mass to find the value of its molar mass

M_M = m/n = "0.627 g"/"0.00394 moles" = color(green)("159 g/mol")#

The answer is rounded to three sig figs.