What is the vapor pressure above a solution prepared by dissolving 0.500 mol of a non-volatile solute in 275 g of hexane (86.18 g/mol) at 49.6°C? P°hexane = 400.0 torr at 49.6°C.

Vapor pressure of solution = 346 Torr

Explanation:

We must work with colligative property of vapor pressure, to solve this:

ΔP = P°. Xm

ΔP is Vapor pressure of pure solvent - Vapor pressure of solution

P° is Vapor pressure of pure solvent

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

Let's find out Xm.

We have the moles of solute, as data → 0.5 moles

Moles of solvent → Mass of solvent / Molar mass of solvent

275 g / 86.18 g/mol = 3.19 moles

Total moles = Moles of solvent + Moles of solute

Xm for solute = 0.5moles / (3.19moles + 0.5 moles) → 0.135

400 Torr - Vapor pressure of solution = 400 Torr. 0.135

Vapor pressure of solution = - (400 Torr. 0.135 - 400 Torr)

Vapor pressure of solution = 346 Torr