The vapour pressure of two pure liquids, A and B, that form an ideal solution are 300 and 800 torr, respectively, at temperature T. A mixture of the vapours of A and B for which the mole fraction of A is 0.25 is slowly compressed at temperature T
(i) What is the composition of first drop of condensate ?
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i)
Sol. pº A = 300 torr pº B = 800 torr
y A = 
= 
= 
x A = 
= 
(ii)
Sol. p = x A pº A + x B pº B = x A pº A + (1 – x A ) pº B
or p = 
× 300 + 
× 800 = 
torr
(iii)
Sol. At normal boiling point, p = 760
p = 760 = x A pº A + (1 – x A ) pº B
or x A = 
= 0.08
(iv)
Sol. Here, the mole fraction of A in the liquid phase will be same as that of A in the original vapour phase x A = 0.25
p = x A pº A + x B pº B = 0.25 × 300 + 0.75 × 800
= 675 torr
(v)
Sol. In this case, we have, y A = 
= 
= 
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