## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605c: 44b

83.9 g KCl

#### Work Step by Step

Molarity M= 0.750 M Volume of solution in litre= 1.50 L Number of moles of KCl n=MV=$0.750M\times1.50L= 1.125 mol$ Mass in grams of KCl= n$\times$ Molar mass of $KCl$ $= 1.125 mol\times74.55 g/mol=83.9 g$

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