## General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal

# Chapter 3 - Chemical Compounds - Example 3-1 - Relating Molar Mass, the Avogadro Constant, and Formula Units of an Iconic Compound - Page 74: Practice Example A

40 g MgCl2

#### Work Step by Step

Since there are 2 chlorine ions in MgCl2, we need $\frac{5.0 \times 10^{23}}{2}$ = $2.5 \times 10^{23}$ fu. $2.5 \times 10^{23}$ fu = $\frac{2.5 \times 10^{23}}{6.0 \times 10^{23}}$ moles. Molecular weight of MgCl2 = 95 g/mol. So mass of MgCl2 = $\frac{2.5 \times 10^{23}}{6.0 \times 10^{23}} \times 95$ = 40 g MgCl2.

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