I don't know how to figure this out.

I think... but I'm not sure... let me give you the problem:

How would you prepare 1000ml of a 5% methenamine solution using a 20% methenamine solution?

I thought about trying the equation with the C1V1 = C2V2. Would that make sense?
Huh?
If you have something that's 20% methenamine and 80% water, and you need something that's 5% methenamine, well, add more water.
If you had pure methenamine and needed 5% solution, it'd be simpler--50mL pure, 950mL water.
As is, you have 1/5th as much methenamine in your methenamine solution as in a bunch of pure methenamine, so you need 5 times as much solution. 250mL 20% solution, 750mL water.
250ml 20% solution, 950




I am not yet born, console me.
I fear that the human race may with tall walls wall me,
with strong drugs dope me, with wise lies lure me,
on black racks rack me, in blood-baths roll me.