Question

How many milliliters of 11.5 M HCl are needed to prepare 830.0 mL of 1.00 M HCI?

Answer 1

`72.174mL`

Explanation:

`M_1V_1=M_2V_2`

`V_1=[M_2V_2]/M_1 = (830times1)/11.5 = 72.174mL`