Question

Determine the approximate freezing point of 0.1 F potassium sulfate solution

Answer 1

$\Delta T_f=$ $i*$ $m*K_f$ ...............(1)

where ΔT_f is the freezing point depression,i is the vanhoff's factor, m is the molality of the solute and K_f is the freezing point depression constant

here value of K_f for water is 1.86 , m is 0.1

now calculating value of i

As potassium sulfate solution will undergo 100 % dissociation

$K_2SO_4\iff2K^++SO_4^{2-}$

so total 3 ions are formed on 100 % dissociation so value of i =3

Substituting these values in the equation (1)

we get $\Delta T_f=3*0.1*1.86=0.558 \ ^0C$

freezing point =$T_f -\Delta T_f$ =$(0-0.558)^0C=-0.558^0C$