In #"1200 g"# of a saturated (#"40 M"#) solution of #"Na"_2"SO"_4# in water at #20^@ "C"#, what is the mass of solute found within it?

NOTE: the given concentration is nonsense and impossible.
- Truong-Son

1 Answer
Truong-Son N.
Aug 11, 2017

Well, apparently you're referring to #"Na"_2"SO"_4#, which has a solubility of less than #"1 M"# at #20^@ "C"#... so you had better check your number.

Straight from Wikipedia, the solubility of anhydrous sodium sulfate is #"13.9 g/100 mL water"# at #20^@ "C"#, or...

#(13.9 cancel"g")/(100 cancel"mL") xx (1000 cancel"mL")/"L" xx "1 mol"/(142.04 cancel"g")#

#= ul"0.979 M"# #"<<"# #"40 M"#!

(In fact, the molarity of PURE water at #25^@ "C"# is #"55.347 M"#, and the solubility of anhydrous #"Na"_2"SO"_4# at #100^@ "C"# is STILL about #"3 M"#... Good luck getting a #"40 M"# solution without vaporizing the water!)

So, in #"1200 g"# of saturated solution at #20^@ "C"#, we have...

#m_"solvent" + m_"solute" = "1200 g"#

We also have (with #rho_(H_2O) = "998.2071 g/L water"# at #20^@ "C"#)

#"13.9 g solute"/(100 cancel"mL solvent") xx cancel"1000 mL"/"998.2071 g water" = "13.9 g solute"/"99.821 g solvent"#,

and thus, #"13.9 g solute"/"113.721 g solution"#.

So, we have a comparison:

#"13.9 g solute"/"113.721 g solution" = m_"solute"/("1200 g solution")#

And so we proportionally have

#ul(m_"solute" = "146.7 g solute")#

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