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

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 13 - Solutions and Their Physical Properties - Exercises - Molarity - Page 593: 22

Answer

$$5.62 \space M \space HNO_3$$

Work Step by Step

1. Calculate the molar mass: $ HNO_3 $ : ( 1.008 $\times$ 1 )+ ( 14.01 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 63.02 g/mol 2. Use the informations as conversion factors to find the molarity of this solution: $$\frac{ 30.00 g \space HNO_3 }{100 \space g \space solution} \times \frac{1 \space mol \space HNO_3 }{ 63.02 \space g \space HNO_3 } \times \frac{ 1.18 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 5.62 \space M$$
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