## Introductory Chemistry (5th Edition)

Published by Pearson

# Chapter 13 - Solutions - Exercises - Problems: 60

#### Answer

(a) 0.069 M (b) 0.0157 M (c) 0.423 M

#### Work Step by Step

Molarity = Number of moles/ Volume (in L) Molarity = (Number of moles X 1000) / Volume (in mL) (a) 1.54 mol of LiCl in 22.2 L of solution = 1.54 / 22.2 = 0.069 M (b) 0.101 mol of $LiNO_{3}$ in 6.4 L of solution = 0.101 / 6.4 = 0.0157 M (c) 0.0323 mol of glucose in 76.2 mL of solution = 0.0323 / 0.0762 = 0.423 M

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