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

2 Answers
Aug 17, 2017

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...


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!

Aug 17, 2017

Answer:

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.