Question

Assume that you have a solution of an unknown solute in cyclohexane. If the solution has a freezing-point depression of 9.50Celcius, what is the molality of this solution? (The molal freezing-point constant of cyclohexane is 20.2 C/m)

Answer 1

We obtain a molality dependent on the assumed van't Hoff factor of `i ~~ 1`,

`m ~~ 0.470` `"molal"`

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We refer to the freezing point depression given by

`DeltaT_f = T_f - T_f^"*" = -iK_fm`,

where:

• `T_f` is the freezing point in `""^@ "C"` of the solution, and `"*"` indicates pure solvent.
• `i` is the van't Hoff factor of the solute, the effective number of dissociated particles per formula unit. For nonelectrolytes, this would be `1`, but we have no idea what this solute is...
• `K_f = 20.2^@ "C/m"` is the freezing point depression constant of cyclohexane at its normal freezing point.
• `m` is the molality in `"mol solute/kg solvent"`.

The molality expression is therefore

`color(blue)(m) = -(DeltaT_f)/(iK_f) = -1/i (-9.50^@ "C")/(20.2^@ "C/m")`

`= color(blue)(0.470/i)` `color(blue)("mol solute/kg solvent")`

So we will have to provide a van't Hoff factor to determine the molality here.

Since this is a fairly high molality (high being higher than `"0.01 molal"`), we have to assume the solute is quite soluble, and thus that it is nonpolar and organic, making it a nonelectrolyte.

That means `i ~~ 1`.

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Of course, had the change in temperature been low enough that `m < "0.01 molal"`, we would have been in the dark about what kind of solute this could have been, as many things are barely soluble in cyclohexane...