Answer
See the answer below.
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:
$m_e=0.107\times 1\ mol\times 46.07\ g/mol=4.93\ g$
$m_w=0.893\times 1\ mol\times 18.015\ g/mol=16.09\ g$
$\chi_e=4.93/(4.93+16.09)\times 100\%=23.5\%$
