Question

How do we prepare a `2.0*mol*L^-1` of `HCl` if `12*mol*L^-1` `HCl` is available...?

Answer 1

Approx. `0.17*L` of acid......... And note that the strongest concentration of `HCl` you can normally get is `10.6*mol*L^-1`.

Explanation:

We need a two litre volume of `1.0*mol*L^-1` `"hydrochloric acid"`.

And this a molar quantity of `2*Lxx1.0*mol*L^-1=2.0*mol`.

And we thus require a volume of `(2*mol)/(12*mol*L^-1)=1/6*L`

Just as an aside, this question has no chemical reality. The conc. laboratory reagent is approx. 33% (w/w); and this translates to a molar concentration of approx. `10.6*mol*L^-1`.

As always with these dilutions, we ADD ACID TO WATER, AND NEVER THE REVERSE.