Chemistry: The Central Science (13th Edition)

Published by Prentice Hall

Chapter 13 - Properties of Solutions - Exercises: 13.51a

Answer

$0.150\;mol\;SrBr_2$

Work Step by Step

Recall that molarity = moles solute / liter solution $\frac{600\;mL\;solution}{}(\frac{1\;L\;solution}{1000\;mL\;solution})(\frac{0.250\;mol\;SrBr_2}{1\;L\;solution})=0.150\;mol\;SrBr_2$

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