# If "6.5 g" of solute X is dissolved in "50 g" of ether solvent ("74 g/mol") whose vapor pressure dropped from "445 mm Hg" to "410 mm Hg", what is the molar mass of X?

Jun 19, 2017

I got $\text{112.88 g/mol}$, though you only have two sig figs. If you use $\text{74 g/mol}$ for the ether molecular weight, you'd get $\text{112.78 g/mol}$.

This is asking you to use Raoult's law to look at vapor pressure reduction from ${P}_{A}^{\text{*}}$ to ${P}_{A}$:

${P}_{A} = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}$

where:

• ${P}_{A}$ is the vapor pressure of solvent $A$.
• $\text{*}$ indicates pure solvent.
• ${\chi}_{A \left(l\right)} = \frac{{n}_{A}}{{n}_{A} + {n}_{X}}$ is the mol fraction of $A$ in the solution phase. ${n}_{A}$ is mols of solvent $A$.

Ether is also called diethyl ether, or ${\text{H"_3"C""H"_2"C"-"O"-"CH"_2"CH}}_{3}$, indeed with a molecular weight of $\text{74.122 g/mol}$. I'll use that, since it's more accurate. Given the mass of solute, we need its mols to find its molar mass.

Those mols can be found from the mol fraction.

${\chi}_{A \left(l\right)} = {P}_{A} / {P}_{A}^{\text{*}}$

$= \left(\text{410 mm Hg")/("445 mm Hg}\right)$

$= {n}_{A} / \left({n}_{A} + {n}_{X}\right) = 0.9213$

We know the mols of the solvent to be:

n_A = m_A/M_A = 50 cancel"g" xx "1 mol ether"/(74.122 cancel"g"),

($m$ being the mass in $\text{g}$ and $M$ being the molar mass in $\text{g/mol}$.)

$\implies {n}_{A} =$ $\text{0.6746 mols ether}$

So, we can solve for the mols of the solute $X$.

$0.9213 = \left(\text{0.6746 mols ether")/("0.6746 mols ether} + {n}_{X}\right)$

$\implies 0.9213 \left(0.6746 + {n}_{X}\right) = 0.6746$

$\implies 0.6215 + 0.9213 {n}_{X} = 0.6746$

$\implies {n}_{X} = \text{0.05758 mols}$

This means the molar mass of $X$ is:

color(blue)(M_X) = m_X/n_X = "6.5 g X"/"0.05758 mols X" = color(blue)("112.88 g/mol")

though you only have two sig figs, apparently.