Question

The vapour pressure of a 5% aqueous solution of a non-volatile organic substance at 373 K is 745 mm Hg. Calculate the molar mass of the substance.

Answer 1

Given: Mass of solute (w₂) = 5 g (since it's a 5% solution in 100 g solution)

Mass of solvent (water) (w₁) = 100 g – 5 g = 95 g = 0.095 kg

Vapour pressure of pure water at 373 K, P° = 760 mm Hg

Vapour pressure of solution, P = 745 mm Hg

M₁ = Molar mass of water = 18 g/mol

Relative lowering of vapour pressure:

`(P^circ - P)/P^circ`

`(760 - 745)/760`

= `15/760`

= 0.0197368

We know that,

`(P^circ - P)/P^circ = (w_2 * M_1)/(w_1 * M_2)`    

`0.0197368 = (5 xx 18)/(95 xx M_2)`

`0.0197368 = 90/(95M_2)`

⇒ 0.0197368 × 95 × M₂​ = 90

⇒ 1.875 × M₂​ = 90

⇒ `M_2 = 90/1.875`

⇒ M₂ = 48 g/mol