Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
The vapour pressures of ethanol and methanol are 42.0 mm and 88.5 mmHg respectively. An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. The mole fraction of methanol in the vapour is:
Q. The vapour pressures of ethanol and methanol are
42.0
mm
and
88.5
mm
H
g
respectively. An ideal solution is formed at the same temperature by mixing
46.0
g
of ethanol with
16.0
g
of methanol. The mole fraction of methanol in the vapour is:
2579
212
Solutions
Report Error
A
0.467
B
0.502
C
0.513
D
0.556
Solution:
P
m
=
P
M
e
O
H
∘
X
M
e
O
H
∘
+
P
E
roH
∘
X
E
r
O
H
Thus
P
M
=
88.5
×
[
32
16
+
46
46
32
16
]
+
42
×
[
32
16
+
46
46
46
46
]
=
57.5
Now
P
M
e
O
H
′
=
P
M
eoH
o
.
X
M
eoH
(
l
)
=
P
m
×
X
M
e
O
H
(
g
)
∴
88.5
×
[
32
16
+
46
46
32
16
]
=
57.5
X
M
eoH
(
g
)
∴
X
M
e
O
H
=
0.513