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14. 0.2 m aqueous solution of KCl freezes at -0.68°C calculate van't Hoff factor. kf, for water is 1.86 K kg mol¹. ​

Question

14. 0.2 m aqueous solution of KCl freezes at -0.68°C calculate van't Hoff factor.

kf k_f for water is 1.86 K kg mol⁻¹.

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Solution

To calculate the van't Hoff factor (i), we can use the formula for the freezing point depression:

ΔTf = i * Kf * m

Where: ΔTf is the freezing point depression, Kf is the cryoscopic constant (also known as the freezing point depression constant), m is the molality of the solution.

We can rearrange this formula to solve for i:

i = ΔTf / (Kf * m)

Given in the problem, we have: ΔTf = 0.68°C (the freezing point depression is the difference between the normal freezing point of the solvent and the freezing point of the solution, since water normally freezes at 0°C, a solution that freezes at -0.68°C has a freezing point depression of 0.68°C), Kf = 1.86 K kg mol⁻¹, m = 0.2 mol/kg.

Substituting these values in:

i = 0.68 / (1.86 * 0.2) = 1.83

So, the van't Hoff factor for this solution is approximately 1.83. This means that each formula unit of KCl in the solution results in about 1.83 particles in the solution.

This problem has been solved

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