Answer
$319\,mL$
Work Step by Step
$M_{2}=\frac{\frac{3.25\,g}{166\,g/mol}}{25.0\times10^{-3}\,L}=0.783\,M$
$M_{1}V_{1}=M_{2}V_{2}$
$\implies V_{2}=\frac{M_{1}V_{1}}{M_{2}}$
$=\frac{5.00\,M\times50.0\,mL}{0.783\,M}$
$=319\,mL$
Introductory Chemistry (5th Edition)
by Tro, Nivaldo J.
Published by
Pearson
ISBN 10:
032191029X
ISBN 13:
978-0-32191-029-5
Chapter 13 - Solutions - Exercises - Cumulative Problems - Page 483: 111
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