Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Questions: 16


We need 40.00 grams of solid NaOH and dissolve it in 1 liter of water. The student would dissolve 0.5 L of 2.00 M NaOH solution and add 0.5 L of water. Three significant figures are necessary to ensure three significant figures in the NaOH molarity.

Work Step by Step

The amount of solid of NaOH we require to form 1.00 L of 1.00 M NaOH solution is: $$1.00 \text{ L} *\frac{1.00 \text{ mol NaOH}}{1.00 \text{ L}}*\frac{40.00 \text{ grams NaOH}}{1\text{ mol NaOH}}=40.00 \text{ grams NaOH}$$ We then add 1 liter of water to these 40.00 grams to create 1.00 L solution of 1.00 M NaOH. We apply the formula: $M_1V_1=M_2V_2$ to calculate the volume of 2.00 M NaOH solution we require: $2.00*V=1.00*1.00\Rightarrow V=0.50$ L. Therefore, we need to combine 0.50 L of 2.00 M NaOH solution with 0.50 L of water to create 1.00 L of 1.00 M solution. By the multiplicative and division rule concerning significant figures when we take the lowest number of significant figures of the factors, both the volume and mass should be determined with at least three significant figures
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