In an ideal binary solution of A and B, when equal mole fraction of A and B are present then total vapour pressure of solution in 760 mm of Hg. If vapour pressure of A in pure state is 600 mm of Hg then vapour pressure of B in pure state will be
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12th
Chemistry
Solutions
Vapour Pressure of Liquid Solutions
Two liquids A and B form an...
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Asked on November 22, 2019 by
Bhavan Dimpy
Two liquids A and B form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of A and 3 mole of B is 550 mm of Hg. At the same temperature, if one more mole of B is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure of A and B in their pure states.
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Let the vapour pressure of pure A be =p
A
0
; and the vapour pressure of pure B be =p
B
0
.
Total vapour pressure of solution (1 mole A + 3 mole B)
=X
A
⋅p
A
0
+X
B
⋅p
B
0
[X
A
is mole fraction of A and X
B
is mole fraction of B]
550=
4
1
p
A
0
+
4
3
p
B
0
2200=p
A
0
+3p
B
0
...(i)
or
Total vapour pressure of solution (1 mole A + 4 mole B)
=
5
1
p
A
0
+
5
4
p
B
0
560=
5
1
p
A
0
+
5
4
p
B
0
2800=p
A
0
+4p
B
0
...(ii)
or
Solving eqs. (i) and (ii),
p
B
0
=600 mm of Hg = vapour pressure of pure B
p
B
0
=400 mm of Hg = vapour pressure of pure
Given:
P°A = 600 mm Hg
P = 760 mm Hg
Mole fraction of A = Mole fraction of B
To Find:
The vapour pressure of B in pure state.
Calculation:
- The mole fraction is same for both A and B
⇒ Mole fraction, XA = XB = 0.5
- Applying Raoult's law, we get:
P = P°A × XA + P°B × XB
⇒ 760 = 600 × 0.5 + P°B ×0.5
⇒ P°B × 0.5 = 760 - 300
⇒ P°B = 460/0.5