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JEE Main Chemistry Class 11 Solutions Part 1 Questions
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© examsnet.com
Question : 62
Total: 100
A solution at
20
∘
C
is composed of
1.5
m
o
l
of benzene and
3.5
m
o
l
of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are
74.7
torr and
22.3
torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively:
[Online April 10,2015]
35.8
torr and
0.280
38.0
torr and
0.589
30.5
torr and
0.389
30.5
torr and
0.480
Validate
Solution:
Total V.P. of solution
=
P
A
∘
X
A
+
P
B
∘
X
B
Given,
P
A
∘
=
74.7
torr,
P
B
∘
=
22.3
torr
n
benzene
=
1.5
mol,
n
toluene
=
3.5
m
o
l
n
solution
=
1.5
+
3.5
=
5
m
o
l
x
A
=
n
benzene
n
solution
=
1.5
5
=
0.3
x
B
=
n
toluene
n
solution
=
3.5
3
=
0.7
Total V.P. of solution
=
(
0.3
×
74.7
+
0.5
×
22.3
)
torr
=
(
22.4
+
15.6
)
torr
=
38
torr
Mole fraction of benzenein vapour form
=
22.4
38
=
0.589
© examsnet.com
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