Answer
1140 mL
Work Step by Step
$M_{2}=\frac{\frac{5.9\,g}{134.45\,g/mol}}{50.0\times10^{-3}\,L}=0.87765\,M$
$M_{1}V_{1}=M_{2}V_{2}$
$\implies V_{2}=\frac{M_{1}V_{1}}{M_{2}}=\frac{8.00\,M\times125\,mL}{0.87765\,M}=1140\,mL$
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: 112
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