## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 4 - Exercises - Page 189: 68

#### Answer

$$1.39 \space M \space ZnCl_2$$

#### Work Step by Step

1. Calculate the amount of moles of $Zn$: $Zn$ : 65.41 g/mol $$25.0 \space g \space Zn \times \frac{1 \space mol \space Zn }{ 65.41 \space g \space Zn } = 0.382\underline{2} \space mol \space Zn$$ 2. Find the amount of moles of $ZnCl_2$ necessary: $$0.382\underline{2} \times \frac{ 1 \space mol \space ZnCl_2 }{ 1 \space mol \space Zn }= 0.382\underline{2} \space mol \space ZnCl_2$$ 3. Find the molarity of $ZnCl_2$: $$Molarity = \frac{ 0.382\underline{2} \space mol \space ZnCl_2 }{ 275 \space mL \space ZnCl_2 } \times \frac{1000 \space mL}{1 \space L} = 1.39 \space M \space ZnCl_2$$

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