Question

The boiling point of an aqueous 1.83 m (nh4)2so4 (molar mass = 132.15 g/mol) solution is 102.5°c. determine the value of the van't hoff factor for this solute if the kb for water is 0.512°c/m. 2.7 1.8 2.3 3.0 3.6

Answer 1

We need to know the value of van't hoff factor.

The van't hoff factor is: 2.66 or 2.7 (approximately)

(NH₄)₂SO₄ is an ionic compound, so it dissociates in solution and produces 3 ionic species. Therefore van't hoff factor is more than one.

From the equation: Δ$T_{b}$=i $K_{b}$.m, where Δ$T_{b}$= elevation of boiling point=102.5 - 100=2.5°C.

m=molality of solute=1.83 m (Given)

$K_{b}$= Ebullioscopic constant or Boiling point elevation constant= 0.512°C/m (Given)

i= Van't Hoff factor

So, 2.5= i X 0.512 X 1.83

i=$\frac{2.5}{0.512 X 1.83}$

i=2.66= 2.7 (approx.)