How do we prepare a #2.0*mol*L^-1# of #HCl# if #12*mol*L^-1# #HCl# is available...?
We need a two litre volume of
And this a molar quantity of
And we thus require a volume of
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.
As always with these dilutions, we ADD ACID TO WATER, AND NEVER THE REVERSE.