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

 0.17 m : Highest freezing point

0.20 m : Second lowest freezing point

0.18 m : 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.

where,

= change in freezing point

i= vant hoff factor

= freezing point constant

m = molality

a) 0.17 m

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

b) 0.18 m

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

c) 0.20 m

For electrolytes undergoing complete dissociation, vant hoff factor is equal to the number of ions it produce. Thus i = 2 for , 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.

answer: c.

explanation: