Question

What volume of concentrated hydrochloric acid (12.0 M `HCl`) is require to make 2.0 liters of a 3.0 M `HCl` solution?

Answer 1

`500*mL` of `12.0*mol*L^-1` `HCl*(aq)` are required.

Explanation:

The product `C_1V_1` is `"concentration"xx"volume"`, and when we multiply the typical units we get `mol*cancel(L^-1)xxcancelL`, i.e. units of `"moles"` as required.

Anyway, as to your problem, `C_1V_1=C_2V_2`, because both sides have the units of moles.

`V_1=(C_2V_2)/C_1=(3.0*mol*L^-1xx2.0*L)/(12.0*mol*L^-1)`

`=1/2*L`

Note that we typically buy conc. `HCl` as a `32%(w/w)` solution, and this is about `10*mol*L^-1`. This question was not well proposed.