Question

How do I prepare a `2.0*mol*L^-1` solution of `"sodium hydroxide"`?

Answer 1

Dissolve approx. `40*g` of `NaOH` in `500*mL` water....

Explanation:

Now sodium hydroxide is deliquescent, i.e. it absorbs water from the air quite rapidly, so be swift when you measure the mass...

And......... `"concentration"="moles of solute"/"Volume of solution"=((40.0*g)/(40.0*g*mol^-1))/(0.500*L)=2.0*mol*L^-1`

Answer 2

You'll need 1 mol of NaOH, because:
`(1mol)/(0.500L)=2.0mol//L=2.0M`

Explanation:

Molar mass of `NaOH=23.0+16.0+1.0=40.0g`
So you weigh out 40 grams of NaOH.

There are two caveats:
(1)
NaOH is very hygroscopic - it absorbs water from the air. It also absorbs carbon dioxide from the air to form soda. In both cases the NaOH will be either diluted or contaminated. If the concentration is to be precise, you may want to titrate against a known concentration of acid.
(2)
Never add the water to the NaOH!
Fill the measuring flask (or whatever) with about 450 mL of water and slowly add the NaOH. Watch the temperature (it will heat up).
After cooling, add water to the 500 mark and mix well.