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"#.

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

See here for further verification.

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...