Why is $0.25 \ \mathrm{mL}$ the correct answer in this problem?
I used the Ksp to find the molarity of Chlorine to be $3.47 \times 10^{-2} \ \mathrm{M}$.
Then I used the formula
$$\text{Molarity}_{\text{initial}} \times \text{Volume}_{\text{initial}} = \text{Molarity}_{\text{final}} \times \text{Volume}_{\text{final}}$$
to find the volume of Chlorine, which turned out to be $0.21 \ \mathrm{mL}$ for me. However, my answer does not match the correct answer.
What am I doing wrong?
First, solve for [$\ce{Cl-}$] using the expression for K$_{\mathrm sp}$ and the given molarity for $\ce{Ag+}$:
$$K_{\mathrm sp} = 1.8\times 10^{-10} = [\ce{Ag+}][\ce{Cl-}]$$
which rearranges to
$$[\ce{Cl-}] = {1.8\times 10^{-10}\over 7.2\times 10^{-5}} = 2.5\times 10^{-6}\, \mathrm{M}$$
Note that the units for K$_{\mathrm sp}$ are actually $\mathrm{M}^{2}$: I have omitted that from the math above but dimensional analysis tells us that the answer for $[\ce{Cl-}]$ is in units of molarity ($\mathrm{M}$).
Because we have 100 $\mathrm{mL}$ of $\ce{Ag+}$ solution, divide the $[\ce{Cl-}]$ molarity by 10 to get the actual number of moles of $\ce{Cl-}$ needed (here, I multiply by 0.1 to achieve the same result):
$$\mathrm{mol}\,\ce{Cl-} = (2.5\times 10^{-6}\,\mathrm{mol}\cdot\mathrm{L}^{-1}) \times 0.1\,\mathrm{L} = 2.5\times 10^{-7}$$
We are given a solution that is 0.001 $\mathrm{M}$ $\ce{Cl-}$, so we can get the volume needed in liters:
$${2.5\times 10^{-7}\,\mathrm{mol}\,\ce{Cl-}\over 10^{-3}\,\mathrm{M}\,\ce{Cl-}} = 2.5\times 10^{-4}\,\mathrm{L}$$
Finally, convert the above quantity to $\mathrm{mL}$: $$\left({1000\,\mathrm{mL}\over\mathrm{L}}\right)\cdot\,2.5\times 10^{-4}\,\mathrm{L} = 0.25\,\mathrm{mL}$$
