Question

What is the concentration of a `40*g` mass of sodium hydroxide dissolved in `250*mL` of water?

Answer 1

Approx. `4*mol*L^-1`.......or `4.0*"normal"`

Explanation:

`[NaOH]=((40*g)/(40*g*mol^-1))/(250xx10^3*mLxx10^-3*L*mL^-1)=4.0*mol*L^-1`

This follows the normal means of calculation, `"concentration"="moles of solute"/"volume of solution"`, and thus get an answer in `mol*L^-1`.

We assume no volume change on dissolution.

And under these circumstances, where the solute causes no further water hydrolysis, `"molarity "-=" normality"`...........