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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