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AIEEE 2010 Solved Paper
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© examsnet.com
Question : 17
Total: 89
On mixing, heptane and octane form an ideal solution. At
373
K
, the vapour pressures of the two liquid components (heptane and octane) are
105
kPa
and
45
kPa
respectively.
Vapour pressure of the solution obtained by mixing
25.0
g
of heptane and
35
g
of octane will be (molar mass of heptane
=
100
g
mol
−
1
and of octane
=
114
g
mol
−
1
)
72.0
kPa
36.1
kPa
96.2
kPa
144.5
kPa
Validate
Solution:
P
Total
=
P
A
∘
x
A
+
P
B
∘
x
B
=
P
∘
Hep
tan
e
x
hep tan
e
+
P
Oc
tan
e
∘
x
O
c
tan
e
=
105
×
25
/
100
25
100
+
35
114
+
45
×
35
/
114
25
100
+
35
114
=
105
×
0.25
0.25
+
0.3
+
45
×
0.3
0.25
+
0.3
=
105
×
0.25
0.55
+
45
×
0.3
0.55
=
26.25
+
13.5
0.55
=
72
kPa
© examsnet.com
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