# What is the freezing point depression when "85.3 g" of oxygen is dissolved in "1500 g" of water? K_f for water at this temperature is 1.86^@ "C/m".

Jul 25, 2017

It would only be $- \text{0.0040"^@ "C}$. The solubility you have quoted is far too high.

See here for further verification.

The solubility of pure ${\text{O}}_{2} \left(g\right)$ in water at ${0}^{\circ} \text{C}$ is about $\text{69.52 mg/L}$... lower than this. We can only take that much, and the rest escapes the water.

At a low solubility like this, the maximum solubility of pure ${\text{O}}_{2} \left(g\right)$ in water at ${0}^{\circ} \text{C}$ is approximately...

(69.52 cancel("mg O"_2))/cancel"L solution" xx cancel"1 L solution"/"1 kg water" xx "1 g"/(1000 cancel"mg")

$\approx$ $\text{0.06952 g/kg water}$

(Here you have $\text{85.3 g"/"1.5 kg}$, or $\text{56.87 g/kg water}$...)

So, the freezing point depression, given by...

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

where ${T}_{f}$ is the freezing point, $\text{*}$ indicates pure solvent, $i$ is the van't Hoff factor, ${K}_{f}$ is given and $m$ is the molality...

is, for nonelectrolytes (where $i = 1$):

color(blue)(DeltaT_f) = -(1)(1.86^@ "C"cdot"kg/mol")((0.06952 cancel"g")/"kg water")("1 mol"/(31.998 cancel("g O"_2)))

$= \textcolor{b l u e}{- {0.0040}^{\circ} \text{C}}$

So you'd see hardly any change. You'd see a boatload of ${\text{O}}_{2}$ escape though... about $\text{85.196 g}$'s worth...