When 10.1 g of an unknown, non-volatile, non-electrolyte, X was dissolved in 100. g of benzene, the vapor pressure of the solvent decreased from 100 torr to 87.7 torr at 299 K. Calculate the molar mass of the solute, X.

56.06 g/mol is the molar mass

Explanation:

Vapor pressure lowering → P° - P' = P°. Xm

Where P° is vapor pressure of pure solvent

P' is vapor pressure of solution

Xm is the mole fraction (moles of solute / total moles)

Total moles = moles of solute + moles of solvent

Let's replace the data.

100 Torr - 87.7 Torr = 100 Torr. Xm

12.3 Torr = 100 Torr. Xm

0.123 = Xm

We know the moles of solvent because we know the molar mass from benzene and its mass in the solution. (mass / molar mass)

100 g / 78 g/mol = 1.28 moles

Let's build the equation where the unknown is the moles of solute

0.123 = moles of solute / moles of solute + 1.28 moles

0.123 (moles of solute + 1.28 moles) = moles of solute

0.123 moles of solute + 0.158 moles = moles of solute

0.158 = 1moles of solute - 0.123moles of solute

0.158 moles = 0.877 moles of solute

0.158 / 0.877 = moles of solute → 0.180

These moles corresponds to 10.1 g of the unknown, non volatile and non electrolyte X compound so:

molar mass (g/mol) → 10.1 g / 0.180 mol = 56.06 g/mol