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A particular unknown solid depresses the freezing point of stearic acid (Kf =4.5 °C/molal) by 3.5 C. If you dissolve the same solid in H2O (Kf = 1.86 C/molal) to the same final molality as you had it in the stearic acid, what would the change in freezing point be?

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The freezing point depression of a solution is directly proportional to the molality of the solute. The freezing point depression of a solution is given by the equation ΔTf = Kf * molality, where ΔTf is the change in freezing point, Kf is the freezing point depression constant of the solvent and molality is the concentration of the solute.

We know that the solid depressed the freezing point of stearic acid (Kf =4.5 °C/molal) by 3.5 °C, and that the same solid was dissolved in water (Kf = 1.86 C/molal) to the same final molality as it had in the stearic acid.

Thus, we can say that the final molality of the solid = 3.5 / 4.5 = 0.777 mol/kg

Now we need to find the freezing point depression for the same solution in water, so we use the freezing point depression constant of water (Kf = 1.86 C/molal) and the final molality of the solid in water

ΔTf = Kf * molality

ΔTf = 1.86 * 0.777 = 1.44 C

So the change in freezing point would be 1.44 C.

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