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NCERT Class XII Chemistry
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Question : 34
Total: 53
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g water.
Solution:  
Given, P0=17.535 mm Hg, w2=25g , w1=450g , Ps=?
For solute (glucose, C6H12O6 ), M2=180gmol–1 ,
For solvent (H2O), M1=18 g mol‌–1
Applying Raoult’s law,
P0−Ps
P0
=
n2
n1+n2

P0−Ps
P0
=
n2
n1
=
w2
M2
w1
M1
 or, 1−
Ps
P0
=
w2M1
w1M2
 (∵n2<<n1)
Substituting the given value, we get
1−
Ps
17.535
=
25×18
450×180
 or, 1−
Ps
17.535
=
1
180

1−
1
180
=
Ps
17.535
 or,
179
180
=
Ps
17.535

or, Ps=17.535×
179
180
 =17.437‌mm‌Hg
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