Question

Which of the following solutes will lower the freezing point of water the most?

Answer 1

Answer: the ionic compound calcium fluoride $(CaF_2)$

Explanation:

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

$T_f$ = change in freezing point

i = Van'T Hoff factor

$k_f$ = freezing point constant

m = molality

1. For $C_{12}H_{22}O_{11}$ , i= 1 as it is a non electrolyte and does not dissociate.

2. For $MgSO_{4}$ , i= 2 as it is a electrolyte and dissociate to give 2 ions.

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

3. For $LiCl$, i= 2 as it is a electrolyte and dissociate to give 2 ions.

$LiCl\rightarrow Li^{+}+Cl^{-}$

4. For $CaF_{2}$, i= 3 as it is a electrolyte and dissociate to give 3 ions.

$CaF_2\rightarrow Ca^{2+}+2F^{-}$

Thus as vant hoff factor is highest factor for $CaF_{2}$ and the freezing point will be lowest.

Answer 2

Freezing point depression is directly proportional to molality. So the compound to have the greatest effect will be the one that disassociates into the most ions = CaF2