# Question 35549

Mar 5, 2015

The volume is $\text{88.5 mL}$.

So, your diluted solution contains 0.100 moles of luminol. The volume of this solution is not important because the number of moles of luminol would be the same regardless of the solution's volume - 1-L, 2-L, 10-L, and so on.

The molarity of the stock solution is defined as the number of moles of solute, in your case luminol, dissolved per liter of solution. This means that every liter of your stock solution will contain 1.13 moles of luminol dissolved.

You can even estimate what the volume will turn out to be; since the number of moles you need from the stock solution to make the diluted one is roughly 10 times smaller than 1.13, a good guess is that you'll need roughly 100 mL of the stock solution.

$C = \frac{n}{V} \implies {V}_{\text{stock") = n_("luminol}} / C$

V_("solution") = "0.100 moles"/"1.13 mol/L" = "0.08849 L"

Rounded to three sig figs and expressed in mL, the answer will be

V_("stock") = "88.5 mL"#