Question

How do we prepare `2*mol*L^-1` `HCl` from `16*mol*L^-1` `HCl`?

Answer 1

We use....approx. `1/8*L` of the stuff........

Explanation:

We use the definition of concentration......

`"Concentration"="Moles of solute"/"Volume of solution"`

Now we want `500*mLxx10^-3*L*mL^-1xx4*mol*L^-1`. Agreed?

And thus here the moles of solute ...........`500*cancel(mL)xx10^-3*L*cancel(mL^-1)xx4*mol*cancel(L^-1)=2*mol`.

We have `16*mol*L^-1` `HCl` available.....and so from the equation above....

`"Volume"="Mole of solute"/"Concentration"=(2*mol)/(16*mol*L^-1)=1/8*L`

`=0.125*L`

Just to add, that this question is chemically unrealistic. The strongest hydrochloric acid you can buy commercially is approx. `"34%(w/w)"`; this has a molar concentration of `12*mol*L^-1`. And practically, when you mix acid and water, you ALWAYS add acid to water.