## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 5 - Exercises - Page 240: 36

420. mL

#### Work Step by Step

$n_{1}$ = 0.553 mol $V_{1}$ = 253 mL $n_{2}$ = (0.553+0.365) mol = 0.918 mol According to Avogadro law, $\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$ Or $V_{2}$ = $\frac{V_{1}\times n_{2}}{n_{1}}= \frac{253mL\times0.918mol}{0.553mol}$ = 420. mL

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