# 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"#.

##### 1 Answer

It would only be

See here for further verification.

The solubility of pure

At a low solubility like this, the **maximum solubility** of pure

#(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

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