Dilution: How many milliliters of `18.4 M H_(2) SO_(4)` are required to prepare `1L` of `0.940 M` solution of `H_(2) SO_(4)`?
Strategy: Since the concentration of the final solution is less than that of the orignal one, this is a dilution process. We are given the molarity of the original solution and the volume `(V_(f))` and molarity `(M_(f))` of the final solution. Thus. Eq (1.21) can be used.

Explanation

Applying the relationship for dillution
`M_(1) V_(1) = M_(f) V_(f)`
We have `V_(1) = (M_(f) V_(f))/(M_(i)) = ((1.00L)(0.940M))/((18.4M))`
`= 0.0510 L = 51.0 mL`