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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

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:

Molefractionofglycerol(solute ,χglycerol)= 0.093Mass ofglycerol = ?Massofwater=425g

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

no.of moles (water) = 425 g 18.0158 g/mol = 23.60 mol

Mole fraction is the 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...

χglycerol = nglycerol nglycerol +  nWater 0.093=nglycerol nglycerol +  23

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