Question

Match the following aqueous solutions with the appropriate letter from the column on the right. Assume complete dissociation of electrolytes. 1. 0.17 m NH4CH3COO ---- A. Lowest freezing point 2. 0.18 m MnSO4 ---- B. Second lowest freezing point 3. 0.20 m CoSO4 ---- C. Third lowest freezing point 4. 0.42 m Ethylene glycol (nonelectrolyte) ---- D. Highest freezing point

Answer 1

Answer:  0.17 m $CH_3COONH_4$ : Highest freezing point

0.20 m $CoSO_4$: Second lowest freezing point

0.18 m $MnSO_4$: Third lowest freezing point

0.42 m ethylene glycol: Lowest freezing point

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,

$\Delta T_f$ = change in freezing point

i= vant hoff factor

$k_f$ = freezing point constant

m = molality

a) 0.17 m $CH_3COONH_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i =2 for $CH_3COONH_4$, thus total concentration will be 0.34 m

b) 0.18 m $MnSO_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for $MnSO_4$, thus total concentration will be 0.36 m

c) 0.20 m $CoSO_4$

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for $CoSO_4$, thus total concentration will be 0.40 m

d) 0.42 m ethylene glycol

For non electrolytes undergoing no dissociation, vant hoff factor is equal to 1 . Thus i = 1 for ethylene glycol, thus concentration will be 0.42 m

The more is the concentration, the highest will be depression in freezing point and thus lowest will be freezing point.