chemistry class

Question

How many milliliters of 11.5 M HCl(aq) are needed to prepare 190.0 mL of 1.00 M HCl(aq)?

Answer 1

V1M1 = V2M2

(x ml)(11.5 M) = (190.0 ml)(1.00 M)

x = 16.52 ml = 16.5 ml (3 sig. figs.)