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 (∆Tf), when 0.01 mol of sodium sulphate is dissolved
in 1 kg of water, is (Kf = 1.86 K kg mol–1)
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 $$\Delta {T_f} = i\,{k_f}.m = 3 \times 1.86 \times 0.01$$
$$ = 0.0558\,K$$
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