Chapter 13, Problem 7PS

### 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

# You wish to prepare an aqueous solution of glycerol, C3H5(OH)3, in which the mole fraction of the solute is 0.093. What mass of glycerol must you add to 425 g of water to make this solution? What is the molality of the solution?

Interpretation Introduction

Interpretation: The mass and molality of glycerol 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...

Molality: The amount of solute (mol) per kilogram of solvent.

Concentration (C, mol/kg) =  molality of solute = Amount of solute (mol)Mass of solvent (Kg)

Weight percent: The mass of one component divided by the total mass of the mixture, multiplied by 100%

weight % A = Mass of A Mass of A + Mass of B + Mass of C + ...×100%

Explanation

Given:

Moleâ€‹â€‰fractionâ€‰ofâ€‰glycerolâ€‰(soluteÂ ,Ï‡glycerol)â€‰=â€‰Â 0.093â€‰MassÂ ofâ€‰glycerolÂ =Â ?Massâ€‰ofâ€‰waterâ€‰â€‰=â€‰â€‰425â€‰g

Number of moles of water can be calculated using mass of water:

no.ofÂ molesÂ (water)Â =Â 425Â gÂ 18.0158Â g/molÂ =Â 23.60Â mol

Mole fraction is the amount of that component divided by the total amount of all of the components of the mixture (Â nAÂ +Â Â nBÂ +Â Â nCÂ +Â ...).

MoleÂ fractionÂ ofÂ AÂ (Ï‡A)=Â â€‰nAÂ nAÂ +Â Â nBÂ +Â Â nCÂ +Â ...

Ï‡glycerolÂ =Â â€‰nglycerolÂ nglycerolÂ +Â Â nWaterÂ 0.093â€‰=â€‰â€‰nglycerolÂ nglycerolÂ +Â Â 23

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