Question

2 liquids X and Y on mixing form an ideal solution.At 30 degree celsiusthe vapour pressure of the solution containing 3 moles of X and 1 mole of Y is 550 mm of Hg.But when 4 moles of X and 1 mole of Y are mixed,the vapour pressure of the solution thus formed is 560 mm of Hg.What would be the vapour pressure of pure X and pure Y at this temperature ?

Answer 1

according to Dalton's law of partial pressure.
$P_{total}=P_X+P_Y$

we also know, partial pressure of each component of the solution is directly proportional to mole fraction of of it.
e.g., $P_x=xP^{\circ}_X$

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
$550mm=xP^{\circ}_X+yP^{\circ}_Y$
550mm = 3/4P°x + 1/4P°y .......(i)

similarly,
case 2 :- mole fraction of X = 4/5
mole fraction of Y = 1/5

now, $560mm=x'P^{\circ}_X+y'P^{\circ}_Y$
560mm = 4/5P°x + 1/5P°y........(ii)

after solving equations (i) and (ii) we get ,
$P^{\circ}_X=600mmHg$
and $P^{\circ}_Y=400mmHg$