Question

What volume of a solution that is `6.0*mol*L^-1` `HCl` is required to make `1.00*L` of a `0.10*mol*L^-1` solution?

Answer 1

We need under `20*mL` of stock solution.

Explanation:

We use the old relationship, `C_1V_1=C_2V_2`. The product in each side gives units of `"moles"`.

And so `6.0*mol*L^-1xxV_1=1.000*Lxx0.1*mol*L^-1`

and thus..............................

`V_1=(1.000*Lxx0.1*mol*L^-1)/(6.0*mol*L^-1)=0.0167*L=16.7*mL.`

Do you add the water to the acid, or the acid to water? (I am not taking the p, the order of addition is important!)