Answer
73.4 kJ
Work Step by Step
Number of moles of $C_{4}H_{10}$
$=\frac{190.\,g}{58.12\,g/mol}=3.27\,mol$
22.44 kJ is needed to vaporize 1 mol of $C_{4}H_{10}$.
$\implies$ Heat transfer required q=
$3.27\,mol\times22.44\,kJ/mol=73.4\,kJ$
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 9 - Liquids, Solids, and Materials - Questions for Review and Thought - Topical Questions - Page 421a: 17
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