JEE MAIN - Chemistry (2010 - No. 23)
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous
solution, the change in freezing point of water (āTf), when 0.01 mol of sodium sulphate is dissolved
in 1 kg of water, is (Kf = 1.86 K kg molā1)
0.0372 K
0.0558 K
0.0744 K
0.0186 K
Explanation
Sodium sulphate dissociates as
$$N{a_2}S{O_4}\left( s \right) \to 2N{a^ + } + SO_4^{ - - }$$
hence van't hoff factor $$i=3$$
Now $$\Delta {T_f} = i\,{k_f}.m = 3 \times 1.86 \times 0.01$$
$$ = 0.0558\,K$$
$$N{a_2}S{O_4}\left( s \right) \to 2N{a^ + } + SO_4^{ - - }$$
hence van't hoff factor $$i=3$$
Now $$\Delta {T_f} = i\,{k_f}.m = 3 \times 1.86 \times 0.01$$
$$ = 0.0558\,K$$