   Chapter 13, Problem 5PS

Chapter
Section
Textbook Problem

What mass of Na2CO3 must you add to 125 g of water to prepare 0.200 m Na2CO3? What is the mole fraction of Na2CO3 in the resulting solution?

Interpretation Introduction

Interpretation: The mass and mole fraction of Na2CO3 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:

MassofNa2CO3=?Massofwater=125=0.125 kgMolalityofNa2CO3=0.200m

Molality of a solute is defined as the amount of solute dissolves in kilogram of solvent.

Molality of solute = Amount of solute (mol)Mass of solvent (Kg)

Molality=MolesofNa2CO3Amountofwater(inKg)0.200m=MolesofNa2CO30.125 kgMolesofNa2CO3=(0.200m)(0.125 kg)=0.025mol

From the number of moles, mass of sodium carbonate can be determined:

MassofNa2CO3=(molesofNa2CO3)(molarmassofNa2CO3)=(0

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