   Chapter 13, Problem 6PS

Chapter
Section
Textbook Problem

What mass of NaNO3 must be added to 500. g of water to prepare a solution that is 0.0512 m in NaNO3? What is the mole fraction of NaNO3 in the solution?

Interpretation Introduction

Interpretation: The mass and mole fraction of NaNO3 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:

MassofNaNO3=?Massofwater=500=0.500 kgMolality=0.0512m

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)

0.0512m=MolesofNaNO30.500 kgMolesofNaNO3=(0.0512m)(0.500 kg)=0.0256mol

Using the number of moles, mass can be determined.

MassofNaNO3=(molesofNaNO3)(molarmassofNaNO3)=(0

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