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TS EAMCET 11-Sep-2020 Shift 1 Solved Paper
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© examsnet.com
Question : 144
Total: 160
The vapour pressure of pure water is 23 mmHg. The vapour pressure of an aqueous solution which contains 10 mass percent of solute ‘A ’ having molecular weight 50 is
0.003 atm
3.45 atm
22 atm
0.028 atm
Validate
Solution:
Given,
Vapour pressure of pure water
=
23
m
m
H
g
10
%
of solute
A
Molecular weight of
A
=
50
Mass of
A
=
10
g
Mass of water
=
100
−
10
=
90
g
Number of moles of
A
=
10
50
=
0.2
Number of moles of water
=
90
18
=
5
∴
Total number of moles
=
5
+
0.2
=
52
Mole fraction of solvent
=
3
5.2
p
s
=
mole fraction of solvent
×
p
∘
5
2.5
×
23
=
22.11
m
m
H
g
≈
22
m
m
H
g
=
22
760
=
0.028
a
t
m
© examsnet.com
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