Question

which of the following aqueous solution has the lowest freezing point. (a)1%glucose. (b)1%sucrose (c)1%Nacl. (d)1%CaCl2

Answer 1

Answer: (d)1%CaCl2

Explanation: Depression in freezing point  is a colligative property which depend upon the amount of the solute.

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

where,

$T_f$ = change in freezing point

i= vant hoff factor

$k_f$ = freezing point constant

m = molality

(a) 1% glucose: For non electrolytes like glucose, vant hoff factor is 1.

(b) 1% sucrose: For non electrolytes like sucrose, vant hoff factor is 1.

(c) 1% $NaCl$ : For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =2 for $NaCl$

(d)1%$CaCl2$: For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =3 for $CaCl_2$.

Thus 1%$CaCl2$ produces most ions and thus lowers the freezing point to maximum.