Two liquids X and Y on mixing from an ideal solution. The vapour pressure of
the solution containing 2 mol of X and 1 mol of Y is 550 mm Hg. But when 4 mol
of X and 1 mole of Y are mixed, the vapour pressure of solution thus formed is 560
mm Hg. What will be the vapour pressure of pure X and pure Y at this temperature?
Answers
Answer:
according to Dalton's law of partial pressure.
P_{total}=P_X+P_YP
total
=P
X
+P
Y
we also know, partial pressure of each component of the solution is directly proportional to mole fraction of of it.
e.g., P_x=xP^{\circ}_XP
x
=xP
X
∘
case 1 :- mole fraction of X = 3/(3 + 1) = 3/4
mole fraction of Y = 1/(3 + 1) = 1/4
we know, Pressure of solution = pressure of X + pressure of Y
550mm=xP^{\circ}_X+yP^{\circ}_Y550mm=xP
X
∘
+yP
Y
∘
550mm = 3/4P°x + 1/4P°y .......(i)
similarly,
case 2 :- mole fraction of X = 4/5
mole fraction of Y = 1/5
now, 560mm=x'P^{\circ}_X+y'P^{\circ}_Y560mm=x
′
P
X
∘
+y
′
P
Y
∘
560mm = 4/5P°x + 1/5P°y........(ii)
after solving equations (i) and (ii) we get ,
P^{\circ}_X=600mmHgP
X
∘
=600mmHg
and P^{\circ}_Y=400mmHgP
Y
∘
=400mmHg
Hope it helps✌