Question

A non-volatile solute is dissolved in methanol with a solubility of $20\;{\text{g}}/100\;{\text{mL}}$. Calculate the molar mass of the solute if the vapour pressure of this saturated solution at ${20^\circ }{\text{C}}$ is $83{\text{mmHg}}$. Given that the density and vapour pressure of methanol are $0.792\;{\text{g}}/{\text{mL}}$ and $95.7{\text{mmHg}}$ at ${20^\circ }{\text{C}}$ respectively.

Answer 1

Hint: To solve this question, we first need to find the mole fraction of the solvent. Then, we need to find the mass of methanol and consequently we will find its number of moles. Later on, we will find the number of moles of the solvent using our Raoult’s law and then finally we will find the molar mass of the solvent as asked in the question.

Formula Used:
We will use the formula of Raoult’s law to solve this question
${P_{{\text{solution }}}} = {X_{{\text{solvent }}}} \times P_{{\text{solvent }}}^ \circ $
Where
 vapour pressure of the solution
${X_{{\text{solvent }}}} = $ mole fraction of the solvent
${{\text{P}}^ \circ }_{{\text{solvent }}} = $ vapour pressure of the pure solvent

Complete step-by-step answer:Now, we will rearrange the formula to get the mole fraction
${X_{solvent}} = \dfrac{{{P_{solution}}}}{{P_{solvent}^ \circ }} = \dfrac{{83{\text{ torr }}}}{{95.7{\text{ torr }}}} = 0.8673$
By the definition of mole fraction, we have
$0.8673 = \dfrac{{{\text{ moles of}} {\text{solvent }}}}{{{\text{ moles}} {\text{of solute + moles}} {\text{of solvent }}}}$
And we already know the solubility which was provided to us in the question $ = \dfrac{{20\;{\text{g solute }}}}{{100\;{\text{mL solution }}}}$
Now, we will find the mass of methanol
${\text{mass}} = 100{\text{mL}} \times 0.792{\text{g/mL}} = 79.2{\text{g}}$
We know that the atomic weight of methanol is \[32{\text{ g}}\]
The number of moles of $79.2{\text{ g}}$ methanol is given as
${\text{moles = }}\dfrac{{79.2\;{\text{g MeOH}}}}{{32 {\text{g methanol}}}} = 2.475{\text{mols}}$
Now, we will use Raoult’s Law
${P_{{\text{solution }}}} = {X_{{\text{solvent }}}} \times P_{{\text{solvent }}}^ \circ $
$83{\text{mm Hg}} = 95.7{\text{mm Hg}} \times {{\text{X}}_{{\text{solvent}}}}$
$X = 0.867 = \dfrac{{{\text{ moles solvent }}}}{{{\text{ moles solvent }} + {\text{ moles solute }}}}$
$0.867 = \dfrac{{2.475{\text{moles}}}}{{(2.475{\text{moles}} + {\text{ moles of}} {\text{solute }})}}$
Moles of solute $ = 0.37$ moles and mass dissolved in the solution is 20g.
Now, the molar mass of the solute is given by
\[\dfrac{{20 {\text{g}}}}{{0.37 {\text{moles}}}} = 54.05 {\text{g mo}}{{\text{l}}^{{\text{ - 1}}}}\]
Hence, the required molar mass is \[54.05 g mo{l^{ - 1}}\]

Note:The law of Raoult states that the partial vapor pressure of a solvent in a solution (or mixture) is equal to or identical to the vapor pressure of the pure solvent multiplied in the solution by its mole fraction.