Question

A solution of 1.25g of a certain non-volatile substance in 20g of water freezes at 271.94K. Calculate the molecular mass of the solute `(K_(f)=1.86" K kg mol"^(-1))`.

Answer 1

Freezing point of solution = 271.94 K
Freezing point of water = 273.0 K
`DeltaT=(273-271.94)=1.06K`
When we that, `m=(1000K_(f)xxw)/(WxxDeltaT)`
Given, `K_(f)=1.86"K kg mol"^(-1),w=1.25g,W=20g andDeltaT=1.06K`
`m=(1000xx1.86xx1.25)/(20xx1.06)=109.66`.