Explanation:
Because molarity is classified as moles of solute per liter of water, dilution of the water may result in a reduction of its concentration.
Therefore, because the amount of moles of solute has to be constant for dilution, you will use the molarity and volume of that same target solution to calculate how many moles of solute will be present in the sample of the stock solution that you dilute.
c = [tex]\frac{n}{v}[/tex]
⇒ n = c ⋅ V
[tex]n_{HCL}[/tex] = 0.250 M ⋅ 6.00 L = 1.5 moles HCl
Now all you have to do is figure out what volume of 6.0 M stock solution will contain 1.5 moles of hydrochloric acid
c = [tex]\frac{n}{v}[/tex]
V = [tex]\frac{n}{c}[/tex]
[tex]V_{Stock}[/tex] = [tex]\frac{1.5 moles}{6.0 \frac{moles}{L} }[/tex] = 0.25 L
Expressed in milliliters, the answer will be
[tex]V_{Stock} = 250ML[/tex] → rounded to two sig figs
