Question

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

Answer 1

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.

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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")`