Question

The vapour pressure of a dilute aqueous solution of a non-volatile solute (molar mass 180 g mol⁻¹) at 373 K is 750 mm Hg. Calculate

1. molality
2. mole fraction of solute.

Answer 1

Given: Vapour pressure of pure water at 373 K (P°) = 760 mm Hg

Vapour pressure of solution (P) = 750 mm Hg

Molar mass of solute = 180 g/mol

By using the Raoult’s law formula for the relative lowering of vapour pressure

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

= `(760 - 750)/760`

= `10/760`

= 0.01316

For a dilute aqueous solution,

`chi_"solute" = n_"solute"/n_"solvent"`

Take 1 kg water as a solvent (so molality will equal moles of solute on this basis).

`n_"solvent" = 1000/18`

= 55.56 mol

`n_"solute" = chi_"solute" xx n_"solvent"`

= 0.01316 × 55.56

= 0.7309 mol

Now, molality is

`m = "moles of ssolute"/"kg of solvent"`

= `0.7309/1`

= 0.731 mol kg⁻¹