Question

A solution prepared by dissolving 0.300 g of an unknown compound in 30 g of CCl₄ has a boiling point that is 0.392°C higher than that of pure CCl₄. What is the molecular weight of solute? (Molal boiling point elevation constant of CCl₄ is 5.02°C/m)

Answer 1

In this case, w = 0.300 g, W = 30 g, ΔT_b = 0.392°C, K_b = 5.02°C/m

Hence, the molecular weight M' of the solute

`M' = (1000 xx K_b xx w)/(W xx Delta T_b)`

= `(1000 xx 5.02 xx 0.300)/(30 xx 0.392)`

= `1506/11.76`

= 128.06 g mol⁻¹