Chapter 13, Problem 6PS

### 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

# 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:

Massâ€‰ofâ€‰NaNO3â€‰=â€‰?Massâ€‰ofâ€‰waterâ€‰=â€‰500â€‰gÂ â€‰=â€‰â€‰0.500Â kgâ€‰â€‰Molalityâ€‰=â€‰0.0512â€‰m

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

Â Â MolalityÂ ofÂ soluteÂ =Â â€‰AmountÂ ofÂ soluteÂ (mol)MassÂ ofÂ solventÂ (Kg)

0.0512â€‰mâ€‰=â€‰â€‰Molesâ€‰ofâ€‰NaNO3â€‰0.500Â kgMolesâ€‰ofâ€‰NaNO3â€‰=(0.0512â€‰m)â€‰â€‰(0.500Â kg)=â€‰0.0256â€‰mol

Using the number of moles, mass can be determined.

Massâ€‰ofâ€‰NaNO3â€‰â€‰=â€‰(molesâ€‰ofâ€‰NaNO3)â€‰â€‰(molarâ€‰massâ€‰ofâ€‰NaNO3)â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰(0

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