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

Aug 17, 2017

Well, molarity is temperature-dependent, so I will assume ${25}^{\circ} \text{C}$ and $\text{1 atm}$...

and I got $\approx$ $\text{1 M}$, because you have only allowed yourself one significant figure...

And at these conditions, ${\rho}_{{H}_{2} O} = \text{0.9970749 g/mL}$, so that

$400 \cancel{\text{g H"_2"O") xx "1 mL"/(0.9970749 cancel"g}}$

$=$ $\text{401.17 mL}$

And so, the molarity is given by:

$\text{M" = "mols solute"/"L solution}$ (and NOT solvent!)

$= \left[\text{20 g NaOH" xx ("1 mol NaOH")/("(22.989 + 15.999 + 1.0079 g) NaOH")]/(401.17 xx 10^(-3) "L solvent" + V_"solute}\right)$

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 ${\text{2.13 g/cm}}^{3}$ of $\text{NaOH}$ at ${25}^{\circ} \text{C}$ to find out its volume contribution.

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

$=$ $\text{0.009390 L}$

And so, the solvent volume will rise by about $\text{9.39 mL}$ (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")

$=$ $\underline{\textcolor{b l u e}{\text{1.218 M}}}$

(Had you assumed V_("soln") ~~ V_"solvent", you would have gotten about $\text{1.25 M}$.)

But you have only allowed yourself one measly significant figure, so I guess you only have a $\text{1 M}$ solution... I guess it'll be inaccurate!

Aug 17, 2017

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.