How can 560.00ml of a 5.6M solution of HCl be made if the only available concentration of HCl is 8.5M?

Answer 1

Mix it with water, that is, if water is available. Otherwise, I'm not sure.

If water is available, you would do it like this:

8.5(x) + 0(560.00-x) = 5.6(560.00)
8.5x = 3136
8.5x/8.5 = 3136/8.5
x = 368.94

So, you need 368.94mL of the 8.5% concentration HCl, and 191.06mL of water (which has a 0% concentration of HCl).

Hope this helps!

Answer 2

It's simple dilution

M1V1=M2V2 =>
V1=(M2/M1)*V2 = (5.6/8.5)*560 =368.9= 369 mL

As a chemist, I assure you that you NEVER prepare such volumes for the dilution of a solution by adding volume of water Vw=Vtotal-Vconcentrated solution.

You ALWAYS take an appropriate volumetric vessel, in your case a 1 L volumetric cylinder, put some water, add the amount of concentrated solution (369 mL of 8.5 M HCl), add water up to the final desired volume (560 mL) and mix.

The ONLY case when you do the approximation that Vfinal=Vsolvent+Vconc.solution is when you are handling extremely small volumes (< 2mL).