Question

Calculate the reduced vapour pressure of chloroform, CHCl3, at 20 ºC after the addition of 0.515 g of the non-volatile solute naphthalene, C10H8, to 60.8 g of chloroform. (The vapour pressure of CHCl3 = 156 mmHg at 20 ºC.)

Answer 1

Calculate the reduced vapour pressure of chloroform, CHCl3, at 20 ºC after the addition of 0.515 g of the non-volatile solute naphthalene, C10H8, to 60.8 g of chloroform. (The vapour pressure of CHCl3 = 156 mmHg at 20 ºC.)

Given:

Grams of CHCl3 = 60.8 g

Grams of C10H8 = 0.515 g

Vapor pressure of CHCl3 = 156 mmHg at 20 ºC.

First, we need to calculate the moles of naphthalene and moles of chloroform:

Napthalene

0.515 g C₁₀H₈ * (1 mol C₁₀H₈ /128.17 g C₁₀H₈) = 0.004018 mol C₁₀H₈

Chloroform

60.8 g CHCl₃ * (1 mol CHCl₃/119.38 g CHCl₃) = 0.50929 mol CHCl₃

The total number of moles is 0.004018 + 0.50929 = 0.5133 mol, and the mole fraction of

chloroform is

Mol fraction CHCl₃ = 0.51929 mol CHCl₃/0.5133 mol = 0.9921

Mol fraction of C₁₀H₈ = 0.04018 mol C₁₀H₈/0.5133 mol = 0.007828

The vapor-pressure lowering is

ΔP = P° X_C10H8 = (156 mmHg)(0.007828) = 1.221 = 1.22 mmHg

Using Raoult's law to calculate the vapor-pressure of chloroform:

P = P°X_CHCl3 = (156 mmHg)(0.9921) = 154.7 mmHg