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JEE Main Chemistry Class 11 Solutions Part 1 Questions
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© examsnet.com
Question : 40
Total: 100
The vapour pressures of pure liquids A and
B
are 400 and
600
m
m
H
g
, respectively at
298
K
. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid
B
is
0.5
in the mixture. The vapour pressure of the final solution, the mole fractions of components
A
and
B
in vapour phase, respectively are:
[April 8,2019(I)]
450
m
m
H
g
,
0.4
,
0.6
500
m
m
H
g
,
0.5
,
0.5
450
m
m
H
g
,
0.5
,
0.5
500
m
m
H
g
,
0.4
,
06
Validate
Solution:
P
=
x
B
p
B
∘
+
x
A
p
A
∘
=
0.5
×
600
+
0.5
×
400
=
300
+
200
=
500
Using the relation
p
i
=
y
i
P
Total
, we can calculate the mole fractions of the components in vapour phase.
p
B
=
y
B
p
total
y
B
=
p
B
P
Total
=
300
500
=
3
5
=
0.6
y
A
=
p
A
P
Total
=
200
500
=
2
5
=
0.4
© examsnet.com
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