Question

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"`.

Answer 1

It would only be `-"0.0040"^@ "C"`. The solubility you have quoted is far too high.

See here for further verification.

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The solubility of pure `"O"_2(g)` in water at `0^@ "C"` is about `"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 `"O"_2(g)` in water at `0^@ "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")`

`~~` `"0.06952 g/kg water"`

(Here you have `"85.3 g"/"1.5 kg"`, or `"56.87 g/kg water"`...)

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

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

where `T_f` is the freezing point, `"*"` 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)))`

`= color(blue)(-0.0040^@ "C")`

So you'd see hardly any change. You'd see a boatload of `"O"_2` escape though... about `"85.196 g"`'s worth...