Question

Which of the following has maximum depression in freezing point. 1. K2So4 2. NaCl 3. Urea 4. Glucose

Answer 1

Depression in freezing point=i×Kf×molality

Answer 2

Answer: $K_2SO_{4}$

Explanation:

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

$\Delta T_f$=  depression in freezing point

i = Van'T Hoff factor  

$k_f$ = freezing point constant

m = molality

1. For $K_2SO_{4}$

, i= 3 as it is a electrolyte and dissociate to give 3 ions.

$K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}$  

2. For $NaCl$

, i= 2 as it is a electrolyte and dissociate to give 2 ions.

$NaCl\rightarrow Na^{+}+Cl^{-}$  

3. For glucose

, i= 1 as it is a non electrolyte and does not dissociate.

4. For urea

, i= 1 as it is a non electrolyte and does not dissociate.

Thus as vant hoff factor is highest for $K_2SO_4$ , thus it has maximum depression in freezing point.