When 8.05 g of an unknown compound X was dissolved in 100. g of benzene, the vapor pressure of the benzene decreased from 100.0 Torr to 94.8 Torr at 26 8C. What is (a) the mole fraction and what is (b) the molar mass of X

Mole fraction for X = 0.052

Molar mass of X = 115.6 g/mol

Explanation:

Lowering vaporpressure → ΔP = P°. Xm

P° is the vapor pressure for pure solvent. → 100 Torr

ΔP = 100 Torr - 94.8 Torr → 5.2 Torr

Let's replace the data given → 5.2 Torr = 100 Torr. Xm

Xm = 5.2 Torr / 100 Torr = 0.052

Xm is the mole fraction for solute so:

Moles of solute / Moles of solute + Moles of solvent = 0.052

We do not know the moles of solute, but we do know the moles of solvent

100 g. 1mol / 78 g = 1.28 moles

Mol of solute / Mol of solute + 1.28 moles = 0.052

Mol of solute = 0.052 mol of solute + 0.066

0.948 mol of solute = 0.066

mol of solute = 0.066 / 0.948 → 0.0696 moles

Now, we can determine the molar mass of X

Molar mass → g/mol → 8.05 g / 0.0696 mol = 115.6 g/mol