Question

What will be the freezing point of a 20.0% solution of glucose, `C_6H_12O_6` in water?

Answer 1

I got `-2.06^@ "C"`.

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I'm assuming `20%` is by mass, not by volume, i.e. it can be written as `"20% w/w"`, or `"20 wt%"`.

Recall the equation for freezing point depression:

`bb(DeltaT_f = T_f - T_f^"*" = -iK_fm_i)`

where:

• `T_f` is the freezing point of the liquid (either solution or solvent), and `""^"*"` indicates/emphasizes that the value is for the pure solvent.
• `i` is the van't Hoff factor, and approximates how many particles per formula unit have dissociated into the solution. For instance, `i = 3` means exactly three ions per solute particle were generated in solution upon dissociation, and that no ion pairing is occurring.
• `K_f = "1.86"^@ "C"cdot"kg"/"mol"` is the freezing point depression constant of water.
• `m_i` is the molality, `"mol solute"/"kg solvent"`, of the solute glucose in the solvent water.
• The negative in front of the far righthand side ensures the molality is positive (as it should be for a physically sensible number).

We can first assume that we have `"100 g"`, or `"0.100 kg"`, of water, so that we know we have `"20 g"` of glucose. That allows us to get a molality of:

`m_i = n_i/w_A = (w_i"/"M_i)/(w_A) = (20 cancel"g glucose"xx "1 mol"/(180.1548 cancel("g glucose")))/("0.100 kg water")`

`= 0.11102/0.100 "mol"/"kg"`

`=` `"1.1102 mol/kg"`

Next, the van't Hoff factor is approximately `i = 1`, since glucose is a non-electrolyte. It dissolves nicely, but does not dissociate well (hardly ionizes at all).

Lastly, we know that `T_f^"*" = 0^@ "C"` for pure water. Therefore, `DeltaT_f = T_f`, and we can solve for `T_f` to get:

`DeltaT_f = -(1)(1.86^@ "C"cdot"kg"/"mol")("1.1102 mol/kg")`

`= color(blue)(-2.06^@ "C" = T_f)`