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

Dec 20, 2016

I got $- {2.06}^{\circ} \text{C}$.

I'm assuming 20% is by mass, not by volume, i.e. it can be written as $\text{20% w/w}$, or $\text{20 wt%}$.

$\boldsymbol{\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = - i {K}_{f} {m}_{i}}$

where:

• ${T}_{f}$ is the freezing point of the liquid (either solution or solvent), and $\text{^"*}$ 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.
• ${K}_{f} = \text{1.86"^@ "C"cdot"kg"/"mol}$ is the freezing point depression constant of water.
• ${m}_{i}$ is the molality, $\text{mol solute"/"kg solvent}$, 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 $\text{100 g}$, or $\text{0.100 kg}$, of water, so that we know we have $\text{20 g}$ of glucose. That allows us to get a molality of:

${m}_{i} = {n}_{i} / {w}_{A} = \left({w}_{i} \text{/"M_i)/(w_A) = (20 cancel"g glucose"xx "1 mol"/(180.1548 cancel("g glucose")))/("0.100 kg water}\right)$

$= \frac{0.11102}{0.100} \text{mol"/"kg}$

$=$ $\text{1.1102 mol/kg}$

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 ${T}_{f}^{\text{*" = 0^@ "C}}$ for pure water. Therefore, $\Delta {T}_{f} = {T}_{f}$, and we can solve for ${T}_{f}$ to get:

$\Delta {T}_{f} = - \left(1\right) \left({1.86}^{\circ} \text{C"cdot"kg"/"mol")("1.1102 mol/kg}\right)$

$= \textcolor{b l u e}{- {2.06}^{\circ} \text{C} = {T}_{f}}$