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Byju's Answer
Standard XII
Chemistry
Dalton's Law of Partial Pressures
The vapour pr...
Question
The vapour pressure of a solution of methanol and ethanol at
20
′
C
was found to be
70
m
m
of Hg. Adding
10
g
of urea to
80
g
of this solution lowers the vapour pressure to
64.6
m
m
of Hg. Determine composition of the original solution
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Solution
P
∘
1
(
m
e
t
h
a
n
o
l
)
,
P
∘
2
(
e
t
h
a
n
o
l
)
P
T
=
P
1
∘
x
1
+
P
∘
2
x
2
=
P
1
∘
(
1
−
x
2
)
+
P
∘
2
x
2
P
T
=
P
1
∘
−
P
1
∘
x
2
+
P
∘
2
x
2
P
T
=
P
1
∘
+
x
(
P
∘
2
−
P
∘
1
)
70
m
m
H
s
=
88.7
+
x
2
(
88.7
−
44.5
)
70
=
88.7
−
44.2
x
2
18.7
=
44.2
x
2
x
2
=
0.42
⇒
E
t
h
a
n
o
l
M
e
t
h
a
n
o
l
x
1
+
x
2
=
1
x
1
=
1
−
0.42
x
1
=
0.58
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0
Similar questions
Q.
The vapour pressure of ethanol and methanol are
44.5
m
m
H
g
and
88.7
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m
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respectively. An ideal solution is formed at the same temperature by mixing
60
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40
g
of methanol. The total vapour pressure of the solution and the mole fraction of methanol in the vapour are respectively:
Q.
The vapour pressure of ethanol and methanol are
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Q.
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g
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respectively. An ideal solution is formed at the same temperature by mixing
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g
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T
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Q.
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Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 2 mole of A and 3 mole of B is 500 mm Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increses by 20 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg);
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