Question

The degree of dissociation of Ca(NO₃)₂ in a dilute aqueous solution containing 7.0 g of the salt per 100 g of water at 100°C is 70 per cent. If the vapour pressure of water at 100°C is 760 mm, calculate the vapour pressure of the solution.

Answer 1

Degree of dissociation of Ca(NO₃)₂ = `70/100 = 0.7`

	\[\ce{Ca(NO3)2 \phantom{..}<=> \phantom{..}Ca^2+ \phantom{.}+ \phantom{..}2NO^-_3}\]
Initially	1 mol                           -                   -
At equilibrium	(1 − 0.7) mol          0.7 mol        2 × 0.7 mol

∴ Total no. of moles in solution = (1 − 0.7) + 0.7 + (2 × 0.7)

= 2.4

∴ `i = "No. of moles in solution"/"No. of moles added"`

= `2.4/1`

= 2.4

Mass of solute = 7.0 g

Normal molecular mass of solute [Ca(NO₃)₂] = 164

∴ Moles of solute = `7/164`

= 0.043

Moles of solvent (H₂O) = `100/18`

= 5.55

∴ Mole fraction of solute `(chi_"solute") = 0.043/(0.043 + 5.55)`

= 7.69 × 10⁻³

Given that vapour pressure of solvent at 100°C = 760 mm

Vapour pressure of solution at 100°C (ρ) = ?

According to the modified equation of relative lowering of vapour pressure, 

`(p^circ - p)/p^circ = i * chi_"solute"`

`(760 - p)/760 = 2.4 xx 7.69 xx 10^-3`

or p = 760 − (760 × 2.4 × 7.69 × 10⁻³)

= 745.97 mm

Hence, the vapour pressure of the given solution at 100°C is 745.97 mm of Hg.