Two liquids A and B from an ideal solution.
At 300k, the v.p of solution containing onemole
of A and 4 mole of B is 560mmHg. At the same
temp. If one mole of B is taken out from the
solution the v.p of the solution has decreased
by 10mmHg, the v.p of Pure A and B are in mm.
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Answer:
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 A
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