Question

Which of the following aqueous solutions will have the lowest freezing point? a. 0.75 m (NH4)3PO4 b. 1.0 m CaSO4 c. 1.0 m LiCl d. 1.5 m CH3OH e. All have the same freezing point.

Answer 1

Answer: a. 0.75m (NH4)3PO4 will have the lowest freezing point .

Explanation: Freezing point decreases as the concentration of the solute substance increases. Assuming the same solvent for all of them, for instance water.

∆T= i.Kf.b

∆T= freezing point depression

i= vant Hoff factor

Kf= molality

Assuming water to be the solvent for all Kf=1.86°C/M

VANT HOFF FACTORS :

For (NH4)3PO4

This has 3 ionic bonding and 1 covalent bonding making it 4 bond

Therefore i=4

For CaSO4

This has 1 metallic bond and 1 covalent bond making it 2 bond.

Therefore i=2

For LiCl

This has 1 metallic bond and 1 non metallic bond making it 2 bond.

Therefore i=2

For CH3OH

This has only 1 covalent bond.

Therefore i=1

MOLALITY:

(NH4)3PO4 = 0.75M

CaSO4= 1.0M

LiCl= 1.0M

CH3OH= 1.5M

FREEZING POINT DEPRESSION:

For (NH4)3PO4

∆T= 4×0.75×1.86=5.58°C

For CaSO4

∆T= 2×1.0×1.86=3.72°C

For LiCl

∆T = 2×1.0×1.86= 3.72°C

For CH3OH

∆T= 1×1.5×1.86=2.79°C

REMEMBER THE HIGHER THE FREEZING POINT DEPRESSION THE LOWER THE FREEZING POINT.

FREEZING POINT DEPRESSION IS THE CHANGE IN THE FREEZING POINT PROPORTIONAL TO THE AMOUNT OF SOLUTE ADDED THE THE SOLUTION.

THEREFORE THE ONE WITH THE LOWEST FREEZING POINT IS (NH4)3PO4

Answer 2

0.75 m (NH4)3PO4 contains  the highest number of particles hence it has the lowest freezing point.

Freezing point is a colligative property. It depends on the amount of substance present in the system. The number of particles present in the system determines the magnitude of the freezing point depression.

We can see that the solution containing 0.75 m (NH4)3PO4 contains four particles. This solution contains the highest number of particles hence it has the lowest freezing point.

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