Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 13 Solutions and Their Behavior - Study Questions - Page 505f: 103


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Work Step by Step

a) In 1 kg of the solution: $n_e=(12/100\times 1000\ g)\div46.07\ g/mol=2.61\ mol$ $n_w=(88/100\times1000\ g)\div18.015\ g/mol=48.85\ mol$ $x_e=2.61/(2.61+48.85)=0.05$ $x_w=1-0.05=0.95$ b) $P_e^0=1\ atm=760\ mmHg$ -> normal boiling point $P_w^0=(78.5-78)/(79-78)\times (341.0-327.3)+327.3=334.15\ mmHg$ -> linear interpolation $P_e=0.05\times 760=38\ mmHg$ $P_w=0.95\times 334.15=317.44\ mmHg$ c) $P=38+317.44=355.44\ mmHg$ $y_e=38/355.44=0.107$ $y_w=317.44/355.44=0.893$ d) Mole fraction increase $0.107/0.05=2.14$ Weight percent: In 1 mole of condensed vapor: $m_e=0.107\times 1\ mol\times 46.07\ g/mol=4.93\ g$ $m_w=0.893\times 1\ mol\times 18.015\ g/mol=16.09\ g$ $\chi_e=4.93/(4.93+16.09)\times 100\%=23.5\%$
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