Question

If a 0.890 m aqueous solution freezes at − 2.40 ∘ C, what is the van't Hoff factor, i , of the solute?

Answer 1

Answer: Van't Hoff factor, i , of the solute is 1.45

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-(-2.40)^0C=2.40^0C$ = Depression in freezing point

i= vant hoff factor = ?

$K_f$ = freezing point constant = $1.86^0C/m$

m= molality =0.890 m

$2.40^0C=i\times 1.86^0C/m\times 0.890$

$i=1.45$

Thus the van't Hoff factor, i , of the solute is 1.45