# Question f294a

Jan 4, 2018

${\text{120 g mol}}^{- 1}$

#### Explanation:

The first thing that you need to do here is to calculate the molality of the solution.

To do that, you can use the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta T = i \cdot {K}_{b} \cdot b}}}$

Here

• $\Delta T$ is the boiling point elevation, measured as the difference between the boiling point of the solution and the boiling point of the pure solvent
• $i$ is the van't Hoff factor, equal to $1$ for non-ionizing solutes
• ${K}_{b}$ is the ebullioscopic constant of the solvent, equal to ${\text{2.65 K kg mol}}^{- 1}$ for benzene $\to$ see here
• $b$ is the molality of the solution

Now, you know that when you dissolve $\text{2.67 g}$ of this unknown solute into $\text{100 g}$ of benzene, you get a boiling point elevation for the resulting solution equal to

$\Delta T = \text{0.531 K}$

Rearrange the equation to solve for $b$.

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

Plug in your values to find

b = (0.531 color(red)(cancel(color(black)("K"))))/(1 * 2.65 color(red)(cancel(color(black)("K"))) "kg mol"^(-1)) = "0.2004 mol kg"^(-1)

This tells you that your solution contains $0.2204$ moles of solute for every $\text{1 kg} = {10}^{3}$ $\text{g}$ of solvent.

You can thus say that your sample will contain

100 color(red)(cancel(color(black)("g benzene"))) * "0.2004 moles solute"/(10^3color(red)(cancel(color(black)("g benzene")))) = "0.02004 moles solute"

Finally, to find the molar mass of the solute, simply divide its mass by the number of moles it contains.

M_ ("M solute") = "2.67 g"/"0.02004 moles" = color(darkgreen)(ul(color(black)("120 g mol"^(-1))))#

I'll leave the answer rounded to two sig figs, but keep in mind that you have one significant figure for the mass of benzene.