Question

A current of dry air was passed through a solution of 2.5 g of a non-volatile solute in 100 g of water and through water alone. The loss in the weight of solution was 1.25 g and that of water was 0.005 g. Calculate the molecular mass of the solute.

Answer 1

Given: Mass of solute (w₂) = 2.5 g

Mass of water solvent (w₁) = 100 g = 0.1 kg

Loss in weight of solution = 1.25 g

Loss in weight of pure water = 0.005 g

Molar mass of water = 18 g/mol

Molar mass of solute (M₂) = ?

`(P^circ - P)/P^circ = ("Loss in wt. of water" - "Loss in wt. of solution")/"Loss in wt. of water"`

= `(1.25 - 0.005)/1.25`

= `1.245/1.25`

= 0.996

So, the relative lowering of vapour pressure is 0.996

Now, `(P^circ - P)/P^circ = (w_2 * M_1)/(w_1 * W_2)` 

`0.996 = (2.5 xx 18)/(100 xx M_2)`

`0.996 = 45/(100 xx M_2)`

⇒ 0.996 × 100 × M₂ = 45

⇒ `M_2 = 45/99.6`

= 113.06 g/mol