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.
for water is 1.86 K kg mol⁻¹.
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.
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