Question

What is the molarity of an aqueous NaOH solution containing 20 g of NaOH and 400 g of water?

Answer 1

Well, molarity is temperature-dependent, so I will assume `25^@ "C"` and `"1 atm"`...

and I got `~~` `"1 M"`, because you have only allowed yourself one significant figure...

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And at these conditions, `rho_(H_2O) = "0.9970749 g/mL"`, so that

`400 cancel("g H"_2"O") xx "1 mL"/(0.9970749 cancel"g")`

`=` `"401.17 mL"`

And so, the molarity is given by:

`"M" = "mols solute"/"L solution"` (and NOT solvent!)

`= ["20 g NaOH" xx ("1 mol NaOH")/("(22.989 + 15.999 + 1.0079 g) NaOH")]/(401.17 xx 10^(-3) "L solvent" + V_"solute")`

We could assume that the solvent volume does not differ from the solution volume, but that is a lie... so let's use the density of `"2.13 g/cm"^3` of `"NaOH"` at `25^@ "C"` to find out its volume contribution.

`20 cancel"g NaOH" xx (cancel"1 mL")/(2.13 cancel"g") xx "1 L"/(1000 cancel"mL")`

`=` `"0.009390 L"`

And so, the solvent volume will rise by about `"9.39 mL"` (about a `2.3%` increase) upon addition of solute. The molarity should then be...

`color(blue)(["NaOH"]) = ["0.5001 mols NaOH"]/(401.17 xx 10^(-3) "L solvent" + "0.009390 L NaOH")`

`=` `ulcolor(blue)("1.218 M")`

(Had you assumed `V_("soln") ~~ V_"solvent"`, you would have gotten about `"1.25 M"`.)

But you have only allowed yourself one measly significant figure, so I guess you only have a `"1 M"` solution... I guess it'll be inaccurate!

Answer 2

1.25M

Explanation:

The other posted solution is detailed and accurate, but possibly "over kill" for this venue. While 20 technically does have only one significant figure without a designated decimal point, I will also assume it to be 20. for calculations.

Also assuming STP for a general chemistry question of this sort, I calculate the moles of NaOH as 0.5 and use the (estimated) density of water at 1g/mL to get 400mL of solvent.

0.5M/400mL = 1.25M solution.

Right on, or certainly within any of the error probabilities, or assumptions of the more "rigorous" answer. You see, chemistry doesn't have to be intimidating.