# You prepare 3.00 L of 0.250 M HNO3 from a stock of nitric acid that is 12.0 M. What volume of the stock solution will you require to make the dilute solution? a) 0.0625 mL b) 62.5 L c) 144 L d) 62.5 mL

Mar 10, 2015

The answer is d) 62.5 mL.

So, you know the volume and the concentration of your target solution, and the concentration of your stock solution. If you use the formula for dilution calculations, you'll have

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where

${C}_{1}$ - the concentration of the stock solution;
${V}_{1}$ - the volume you need to take from the stock solution;
${C}_{2}$ - the concentration of your target solution;
${V}_{2}$ - the volume of the target solution.

Plug all your values into the above equation and solve for ${V}_{1}$

${V}_{1} = {C}_{2} / {C}_{1} \cdot {V}_{2} = \text{0.250 M"/"12.0 M" * "3.00 L}$

${V}_{1} = \text{0.0625 L}$

Expressed in mililiters, the answer will be

$\text{0.0625 L" * "1000 mL"/"1 L" = "62.5 mL}$