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

The answer to your question is: V1 = 74.35 ml

Step-by-step 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