Question

What will be the vapour pressure of a 1 molal aqueous solution of MgCl₂, assuming dissociation of MgCl₂ to be 70 mole percent?

(Vapor pressure of pure water at 25°C is 23.8 mm Hg.)

Answer 1

`P_A^°` = 23.8 mm Hg

m = 1 molal, 1 mol of solute in 1000 g of water

n_B = 1 mol

n_A = `1000/18`

= 55.5 mol

\[\ce{MgCl2 -> Mg^2+ + 2Cl^-}\]
1	0	0
1 − 0.7	0.7	1.4

n = 3

i = 1 + α(n − 1)

= 1 + 0.7(3 − 1)

= 1 + 0.7 × 2

= 1 + 1.4

= 2.4

Using modified Raoult’s law:

`(P_A^° - P_s)/P_A^° = i((n_B)/(n_A + n_B))`

`(23.8 - P_s)/23.8 = 2.4((1)/(55.5 + 1))`

`(23.8 - P_s)/23.8 = 2.4((1)/(56.5))`

`(23.8 - P_s)/23.8 = 2.4(0.0177)`

`(23.8 - P_s)/23.8 = 0.04248`

23.8 − P_s = 23.8 × 0.04248

23.8 − P_s = 1.011024

P_s = 23.8 − 1.011024

P_s = 22.788

P_s ≈ 22.8 mm Hg