# What is the freezing point depression of a solution containing 100g of ethanol (C_2H_5OH) in 0.750 kg of water?

Jan 2, 2018

Well, I got $- {5.38}^{\circ} \text{C}$, but I only get 1 significant figure, so... $- {5}^{\circ} \text{C}$.

The change in freezing point due to adding solute into a solvent is given by:

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

where:

• ${T}_{f}$ is the freezing point of the solvent in the context of the solution. $\text{*}$ indicates pure solvent instead.
• $i$ is the van't Hoff factor, i.e. the effective number of solute particles per undissociated particle placed into the solvent.
• ${K}_{f} = {1.86}^{\circ} \text{C"cdot"kg/mol}$ is the freezing point depression constant of water.
• $m$ is the molality of the solution, or the $\text{mols solute/kg solvent}$.

Ethanol dissolves well in water, but does not dissociate much at all, since their ${\text{pK}}_{a}$ values are nearly identical. Thus, $i \approx 1$.

The mols of ethanol in solution are:

$100 \cancel{\text{g EtOH" xx ("1 mol EtOH")/(46.0684 cancel"g EtOH") = "2.17 mols}}$

So, the molality is:

$m = \text{2.17 mols EtOH"/"0.750 kg water}$

$=$ $\text{2.89 mol/kg}$

The change in freezing point is then:

$\Delta {T}_{f} = - 1 \cdot {1.86}^{\circ} \text{C"cdotcancel"kg/mol" cdot 2.89 cancel"mol/kg}$

$= - {5.38}^{\circ} \text{C}$

But since you only provided 1 significant figure, we can only say...

$\textcolor{b l u e}{\Delta {T}_{f} \approx - {5}^{\circ} \text{C}}$