Chemistry: The Central Science (13th Edition)

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

Chapter 4 - Reactions in Aqueous Solution - Exercises: 4.62a

Answer

Approximately 0.1029M.

Work Step by Step

1. Find the molar mass of $Na_2CrO_4:$ Na: 23 * 2 = 46 Cr: 52 * 1 = 52 O: 16 * 4 = 64 46 + 52 + 64 = 162g/mol 2. Find the number of mols in 12.5g of $Na_2CrO_4$: $mm = \frac{mass(g)}{n(moles)} $ $162 = \frac{12.5}{n(moles)}$ $n(moles) = \frac{12.5}{162}$ $n(moles) \approx 0.07716$ 3. Now, find the concentration: $C = \frac{n(moles)}{V(L)}$ $C = \frac{0.07716}{0.750}$ $C \approx 0.10288M $
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