Question

How many milliliters of 11.5 M HCl(aq) are needed to prepare 855.0 mL of 1.00 M HCl(aq)?

Answer 1

Answer:

The answer to your question is:   V1 = 74.35 ml

Explanation:

Data

Volume 1 = V1 = ?

Concentration 1 = C1 = 11.5 M

Volume 2 = V2 = 855 ml

Concentration  2 = 1 M

Formula

                            C1V1 = C2V2

                           V1 = C2V2 / C1

                           

Process

                          V1 = (1 x 855) / 11.5

                          V1 = 855 / 11.5

                          V1 = 74.35 ml