   Chapter 13, Problem 32PS

Chapter
Section
Textbook Problem

A solution of glycerol, C3H5(OH)3, in 735 g of water has a boiling point of 104.4 °C at a pressure of 760 mm Hg. What is the mass of glycerol in the solution? What is the mole fraction of the solute?

Interpretation Introduction

Interpretation: The mass and mole fraction of glycerol has to be identified.

Concept introduction:

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
• The Mole fraction:

Given data:

Massofwater=735=0.735 kgBoiling point of solution  = 104.4oCPressure =760mmHg

(ΔTbp)  = 104.4oC - 100oC = 4.4oCKbpofWater=+0.5121oC/m

Elevation in boiling point = ΔTbp= Kbp. msolute,4.4oC = (+0.5121oC/m). msolutemsolute=4.4oC(+0.5121oC/m)=8.59m

Molality=MolesofglycerolAmountofwater(inkg)

8.59m=Molesofglycerol0.735 kgMolesofglycerol=(8.59m)(0.735 kg)=6

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