   Chapter 13, Problem 75GQ

Chapter
Section
Textbook Problem

If a volatile solute is added to a volatile solvent, both substances contribute to the vapor pressure over the solution. Assuming an ideal solution, the vapor pressure of each is given by Raoult’s law, and the total vapor pressure is the sum of the vapor pressures for each component. A solution, assumed to be ideal, is made from 1.0 mol of toluene (C6H5CH3) and 2.0 mol of benzene (C6H6). The vapor pressures of the pure solvents are 22 mm Hg and 75 mm Hg, respectively, at 20 °C. What is the total vapor pressure of the mixture? What is the mole fraction of each component in the liquid and in the vapor?

Interpretation Introduction

Interpretation: Mole fraction of toluene and benzene in liquid as well as in vapor has to be determined. Vapor pressure of the mixture has to be calculated.

Concept introduction:

Raoult’s law: In a solution, vapor pressure of solvent is proportional to its mole fraction.

Psolvent=XsolventP0solvent

where,

P0solvent is the vapor pressure of pure solvent.

Mole fraction: Amount of that component divided by the total amount of all of the components of the mixture ( nA +  nB +  n...).

Mole fraction of A (χA)= nA nA +  nB +  n...

Explanation

Mole fraction of toluene and benzene in liquid as well as vapour and also the vapour pressure of the mixture was calculated.

Given,

Amountoftoluene,ntoluene=1.0molAmountofbenzene,nbenzene=2.0molPartialpressureoftoluene,P0toluene=22mmHgPartialpressureofbenzene,P0benzene=75mmHg

• Mole fraction of toluene and benzene in liquid is calculated

Mole fraction of toluene (χtoluene)= ntoluene ntoluene +  nbenzene=1mol(1mol+2mol)=0.333

Mole fraction of toluene in liquid is 0.333

Mole fraction of benzene (χbenzene)= nbenzene ntoluene +  nbenzene=2mol(1mol+2mol)=0.667

Mole fraction of benzene in liquid is 0

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