AP EAMCET Engineering 2013 Solved Paper

According to Raoult’s law

p°−ps

p°

=

n2

n1+n2

where, p°= vapour pressure of pure water at 100°C
= 760 mmHg.
ps= vapour pressure of solution at 100°C
n2= moles of solute =

w2

M2

=

18

180

=0.1mol
n1= moles of solvent =

w1

M1

=

180

18

=10mol
By putting these values in the formula

p°−ps

p°

=

0.1

10+0.1

or 10.1(p°−ps)=0.1p°
or 10p°=10.1ps
or ps=

10×760

10.1

=752.4mmHg