Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 11 - Reactions in Aqueous Solutions II: Calculations - Exercises - Molarity - Page 393: 3

Answer

a) $0.5 \frac{mol}{dm^{3}}$ b) $0.1543 \frac{mol}{dm^{3}}$ c) $0.0873 \frac{mol}{dm^{3}}$

Work Step by Step

a) $m(H_{3}AsO_{4})=35.5g$ $V=500ml=0.5dm^{3}$ $c=\frac{n(H_{3}AsO_{4})}{V} = \frac{\frac{m(H_{3}AsO_{4})}{M(H_{3}AsO_{4})}}{V} = \frac{m(H_{3}AsO_{4})}{M(H_{3}AsO_{4})V} = \frac{35.5 g}{142 \frac{g}{mol}\times0.5dm^{3}}=0.5\frac{mol}{dm^{3}}$ b) $m((COOH)_{2})=8.33g$ $V=ml=0.6dm^{3}$ $c=\frac{m((COOH)_{2})}{M((COOH)_{2})V} = \frac{8.33 g}{90 \frac{g}{mol}\times0.6dm^{3}}=0.1543\frac{mol}{dm^{3}}$ c) $m((COOH)_{2}\times2H_{2}O)=8.25g$ $V=750ml=0.75dm^{3}$ By dissolving crystalohydrate in water, the following reaction occurs: $(COOH)_{2}\times2H_{2}O\rightarrow(COOH)_{2}+2H_{2}O$ The amount of $(COOH)_{2}$ obtained is equal to the initial amount of $(COOH)_{2}\times2H_{2}O$ (since matching coefficients in the reaction are in $1:1$ ratio). $n((COOH)_{2})=n((COOH)_{2}\times2H_{2}O)=\frac{8.25g}{126\frac{g}{mol}}=0.0655mol$ The amount of water obtained by dissolving this crystalohydrate is negligible, as compared to the volume of $V=750ml$. Hence, $V_{total} = 750ml$. $\implies c=\frac{n((COOH)_{2})}{V_{total}}=\frac{0.0655mol}{0.75dm^{3}}=0.0873\frac{mol}{dm^{3}}$.
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