Question

Solution is prepared by dissolving 4.57 g magnesium chloride in 43.24 g of water . The vapour pressure above the solution at 348 K is
found to be 275.5 torr. The vapour pressure of pure water at same temperature is 289.1 torr. Calculate the percentage dissociation of magnesium chloride . What is the Van’t Hoff’s factor ?

Answer 1

P• = 289.1 torr
P = 275.5 = (Xi)P•
Xi = mile fraction of liquid

Xi = P/P•
= 275.5/289.1
= 0.953(approx)

moles of water = 43.24/18 = 2.402
moles of MgCl2 = 4.57/95 = 0.048(approx)

for moles of dissociation of MgCl2
Xi = 2.4/(2.4+n)
0.95 = 2.4/(2.4+n)
n=0.1mole

MgCl2 --> Mg+² + 2Cl-
if x moles of MgCl2 dissociate then x moles of mg and 2x moles of cl will formed

(0.05-x)+3x = 0.1
x = 0.025
%dissociation of MgCl2 = 0.025/0.05×100 = 50%