Question

If 0.1 m aqueous solution of calcium phosphate is 80% dissociated then the freezing point of the solution will be (kf of water = 1.86 K kg mol-¹

Answer 1

dissociation of calcium phosphate is ...

$Ca_3(PO_4)_2\Leftrightarrow 3Ca^2++2PO_4^3-$

number of ions , n = 5

so, Van't Hoff factor , i = $1-\alpha+n\alpha$

given, degree of dissociation, $\alpha$ = 0.8

so, i = 1 - 0.8 + 5 × 0.8 = 1 - 0.8 + 4 = 4.2

now, freezing point , Tf = iKfm

where m is molality.

here, Kf = 1.86 K kg/mol , m = 0.1

so, freezing point , Tf = 4.2 × 1.86 × 0.1

= 0.7812 K

hence, freezing point of the solution will be 0.7812 K