Question

What is the change in freezing point when `"54 g"` of glucose (use `"FW" = "180 g/mol"`) is dissolved in `"0.250 kg"` of water? `K_f = 1.86^@ "C/m"` for water.

Answer 1

`-2.23^@ "C"`

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Well, the fact is, we (usually) don't care what the solute is. All we needed was how many `"mols"` of it we have (which we can get regardless of what its name is), whether it's a strong electrolyte or not, and how many `"kg"` of water there are.

The decrease in freezing point is given by:

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

where:

• `T_f` is the freezing point of the solution.
• `T_f^"*"` is the freezing point of the pure solvent.
• `i` is the van't Hoff factor of the solute, i.e. the number of particles per formula unit that have gone into solution upon any dissociation.
• `K_f = 1.86^@ "C/m"`, or `""^@ "C"cdot"kg/mol"` is the freezing point depression constant of water.
• `m` is the molality of the solution in `"mol solute/kg solvent"`.

Given the mass of the glucose, we have:

`54 cancel"g solute" xx "1 mol"/(180 cancel"g glucose")`

`=` `"0.30 mols glucose"`

And we know that `"250 g"` is `"0.250 kg"` of water. Water is, of course, the solvent. So:

`m = "0.30 mols glucose"/"0.250 kg water"`

`=` `"1.20 mol/kg"`

Therefore, noting that glucose is a NONelectrolyte (i.e. its van't Hoff factor is `1`), the change in freezing point is:

`DeltaT_f = -(1)(1.86^@ "C"cdot"kg/mol")("1.20 mol/kg")`

`= -2.23^@ "C"`

Therefore, if we wished, we could easily find the new freezing point.

`T_f - T_f^"*" = DeltaT_f`

`=> color(blue)(T_f) = DeltaT_f + cancel(T_f^"*")^(0^@ "C")`

`= color(blue)(-2.23^@ "C")`

Naturally, it's the same value as the change in freezing point, since the freezing point of water is `0^@ "C"` at `"1 atm"`.