A solution containing 30 g of non-volatile solute exactly in 90 g of water has a
vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the
solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
(i)molar mass of the solute (ii) Vapour pressure of water at 298 K
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Weight of solute, W
B
=30 g,
Weight of water, W
A
=90 g
Vapour pressure of solution P
A
=2.8 kPa
According to Raoult's law,
P
A
o
P
A
o
−P
A
=X
B
≃
M
B
W
A
W
B
M
A
P
A
o
P
A
o
−2.8
=
M
B
×90
30×18
P
A
o
2.8
=
M
B
M
B
−6
......(1)
After adding water-
Weight of solute W
B
=30 g, Weight of water W
A
=90+18=108 g
Vapour pressure of solution P
A
=2.9 kPa
According to Raoult's law,
P
A
o
P
A
o
−P
A
=X
B
≃
M
B
W
A
W
B
M
A
P
A
o
P
A
o
−2.9
=
M
B
×108
30×18
P
A
o
2.9
=
M
B
M
B
−5
......(2)
Divide equation (1) by equation (2), we get
2.9
2.8
=
M
B
−5
M
B
−6
M
B
=34 g/mol
Substituting the values of M
B
in equation (1), we get
P
A
o
2.8
=
34
34−6
P
A
0
=3.4 kPa
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