#### Answer

83.9 g KCl

#### Work Step by Step

Molarity M= 0.750 M
Volume of solution in litre= 1.50 L
Number of moles of KCl n=MV=$0.750M\times1.50L= 1.125 mol$
Mass in grams of KCl= n$\times$ Molar mass of $KCl$
$= 1.125 mol\times74.55 g/mol=83.9 g$

Published by
Cengage Learning

ISBN 10:
1285199049

ISBN 13:
978-1-28519-904-7

83.9 g KCl

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