# 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

Aug 11, 2017

Well, apparently you're referring to ${\text{Na"_2"SO}}_{4}$, which has a solubility of less than $\text{1 M}$ at ${20}^{\circ} \text{C}$... so you had better check your number.

Straight from Wikipedia, the solubility of anhydrous sodium sulfate is $\text{13.9 g/100 mL water}$ at ${20}^{\circ} \text{C}$, or...

$\left(13.9 \cancel{\text{g")/(100 cancel"mL") xx (1000 cancel"mL")/"L" xx "1 mol"/(142.04 cancel"g}}\right)$

$= \underline{\text{0.979 M}}$ $\text{<<}$ $\text{40 M}$!

(In fact, the molarity of PURE water at ${25}^{\circ} \text{C}$ is $\text{55.347 M}$, and the solubility of anhydrous ${\text{Na"_2"SO}}_{4}$ at ${100}^{\circ} \text{C}$ is STILL about $\text{3 M}$... Good luck getting a $\text{40 M}$ solution without vaporizing the water!)

So, in $\text{1200 g}$ of saturated solution at ${20}^{\circ} \text{C}$, we have...

${m}_{\text{solvent" + m_"solute" = "1200 g}}$

We also have (with ${\rho}_{{H}_{2} O} = \text{998.2071 g/L water}$ at ${20}^{\circ} \text{C}$)

$\text{13.9 g solute"/(100 cancel"mL solvent") xx cancel"1000 mL"/"998.2071 g water" = "13.9 g solute"/"99.821 g solvent}$,

and thus, $\text{13.9 g solute"/"113.721 g solution}$.

So, we have a comparison:

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

And so we proportionally have

$\underline{{m}_{\text{solute" = "146.7 g solute}}}$