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?

1 Answer
Jun 1, 2017

Answer:

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!)