Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 13 - Properties of Solutions - Exercises - Page 569: 13.57b



Work Step by Step

For ease of calculation, we will assume that there is 100 g of sample. Recall that molarity is mol solute / L solution. First, find moles of Zn present (molar mass = 65.38 g/mol). $100\;g\times20.0\%\;Zn=20.0\;g\;Zn$ $\frac{20.0\;g}{}(\frac{1\;mol}{65.38\;g})=0.306\;mol\;Zn$ Next, find volume of solution using the given density: $8750\;kg/m^3=8750\;kg/(1000\;L)=8.750\;kg/L$ $\frac{100\;g\;sample}{}(\frac{1\;kg}{1000\;g})(\frac{1\;L}{8.750\;kg})=0.0114\;L\;solution$ Next, calculate molarity: $\frac{0.306\;mol\;Zn}{0.0114\;L\;solution}=26.8\;M\;Zn$
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