Chapter 13, Problem 32PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
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:

Massâ€‰ofâ€‰waterâ€‰=â€‰735â€‰gÂ â€‰=â€‰â€‰0.735Â kgBoilingÂ pointÂ ofÂ solutionÂ Â =Â â€‰104.4oCPressureÂ =â€‰760mmHg

(Î”Tbp)Â Â =Â â€‰104.4oCÂ -Â 100oCÂ =Â 4.4oCKbpofâ€‰Water=â€‰+0.5121oC/m

Elevationâ€‰Â inÂ boilingÂ pointÂ =Â Î”Tbp=Â Kbp.Â msolute,4.4oCÂ =Â (+0.5121oC/m).Â msolutemsolute=â€‰4.4oC(+0.5121oC/m)=â€‰8.59â€‰m

Molalityâ€‰=Molesâ€‰ofâ€‰glycerolAmountâ€‰ofâ€‰waterâ€‰(inâ€‰kg)

8.59â€‰mâ€‰=â€‰â€‰Molesâ€‰ofâ€‰glycerolâ€‰0.735Â kgMolesâ€‰ofâ€‰glycerolâ€‰=(8.59â€‰m)â€‰â€‰(0.735Â kg)=â€‰6

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