   Chapter 13, Problem 76GQ

Chapter
Section
Textbook Problem

A solution is made by adding 50.0 mL of ethanol (C2H5OH, d = 0.789 g/mL) to 50.0 mL of water (d = 0.998 g/mL). What is the total vapor pressure over the solution at 20 °C? (See Study Question 75.) The vapor pressure of ethanol at 20 °C is 43.6 mm Hg.

Interpretation Introduction

Interpretation: Total vapor pressure of the solution at 200C has to be calculated if the vapor pressure of ethanol is 43.6mmHg

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...

Density is calculated by the equation,

Density=massvolume

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Explanation

Total vapour pressure of the solution at 200C is calculated.

Given,

Densityofethanol=0.789g/mLDensityofwater=0.998g/mLPartialpressureofethanol,P0ethanol=43.6mmHgPartialpressureofwater,P0water=17.5mmHg

Density is calculated by the equation,

Density=massvolume

Mass is calculated,

Massofwater=density×volume=0.998g/mL×50mL=49.9g

Mass of water is 49.9g

Massofethanol=density×volume=0.789g/mL×50mL=39.45g

Mass of ethanol is 39.45g

The number of moles of any substance can be determined using the equation

Numberofmole=GivenmassofthesubstanceMolarmass

Number of moles of water=49.9g18.02g/mol=2.77mol

Number of moles of water is 2.77mol

Number of moles of ethanol=39.45g46.07g/mol=0.856mol

Number of moles of ethanol is 0.856mol

Mole fraction of ethanol and water in liquid is calculated

Mole fraction of ethanol (χethanol)= nethanol nethanol +  nwater=0

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