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

Step-by-step explanation:

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

`\Delta T_f=i* k_f* 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.