A solution containing 8.3 g of a nonvolatile, nondissociating substance dissolved in 1.00 mol of chloroform, CHCl3, has a vapor pressure of 511 torr. The vapor pressure of pure CHCl3 at the same temperature is 526 torr. Calculate (a) the mole fraction of the solute, (b) the number of moles of solute in the solution, and (c) the molecular mass of the solute.

a. Xm = 0.0229

b. 0.0234 moles

c. 354.1 g/mol

Explanation:

ΔP = P° . Xm

ΔP = P° - P', where P° is vapor pressure of pure solvent and P', vapor pressure of solution-

This is the formula for lowering vapor pressure.

If we apply the data given: 523 Torr - 511 Torr = 523 . Xm

Xm = ( 523 Torr - 511 Torr) / 523 Torr → 0.0229

Xm = Mole fraction of solute → Moles of solute / Total moles (sv + solute)

We can make this equation to determine moles of solute

0.0229 = Moles of solute / Moles of solute + 1

0.0229 (Moles of solute + 1) = Moles of solute

0.0229 = Moles of solute - 0.0229 moles of solute

0.0229 = 0.9771 moles of solute → 0.0229 / 0.9971 = 0.0234 moles

Molecular mass of solute → g/mol → 8.3 g / 0.0234 mol = 354.1 g/mol