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

$72.174 m L$
${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$
${V}_{1} = \frac{{M}_{2} {V}_{2}}{M} _ 1 = \frac{830 \times 1}{11.5} = 72.174 m L$