Scoring: Your score will be based on the number of correct matches minus the number of incorrect matches. There is no penalty for missing matches.
A) Match the solutions to the descriptions of the freezing points.
Solutions:
1. One mole of the jonic compound Na3PO4 dissolved in 1000 g of H2O
2. One mole of the ionic compound CuSO4 dissolved in 1000 g H2O
3. One mole of the nonelectrolyte C6H12O6 dissolved in 1000 g H2O
Descriptions:
O lowest freezing point
O highest freezing point
O intermediate freezing point

One mole of the jonic compound dissolved in 1000 g of → lowest freezing point

One mole of the jonic compound dissolved in 1000 g of → intermediate freezing point

One mole of the jonic compound dissolved in 1000 g of → highest freezing point

Explanation:

where,

=depression in freezing point =  

= freezing point constant  

m = molality

i = van't Hoff factor

= freezing point of solution

T = freezing point of pure solvent

Since the moles of solute and mass of solvent is same in all cases.So is directly related to and

Higher the value of 'i' more will be depression in freezing and lowest will be freezing point of the solution.

1) One mole of the jonic compound dissolved in 1000 g of

i = 4

This solution will have the lowest freezing point.

2) One mole of the jonic compound dissolved in 1000 g of

i = 2

This solution will have the intermediate  freezing point.

3)One mole of the jonic compound dissolved in 1000 g of

Glucose is a non electrolyte substance.

i = 1

This solution will have the highest freezing point.

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