Question

Describe a different set of dilutions to produce 3 mg Fe/mL from 1 000 mg

Fe/mL

Short answer

Dilute 10 mL of stock to 100 mL to get$\left[\mathrm{Fe}\right]=100\mathrm{\mu gFe}/\mathrm{mL}$. Dilute $3\mathrm{x}1.00\mathrm{mL}$of the new solution up to 100 mL to get$3\mathrm{\mu gFe}/\mathrm{mL}$

Step-by-step solution

Calculate concentration:

$\mathrm{Concentration}\mathrm{of}\mathrm{Fe}=\frac{(\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{diluted})}{\mathrm{Volume}\mathrm{of}\mathrm{stock}\mathrm{solution}}(\mathrm{Initial}\mathrm{concentration}\mathrm{of}\mathrm{Fe})$

Given,

Volume of solution diluted =$10\mathrm{mL}$

Volume of stock solution =$100\mathrm{mL}$

Initial concentration of Fe =$1000\mathrm{\mu g}/\mathrm{mL}$

Substitute in above equation,

Concentration of Fe =$\left(\frac{10\mathrm{mL}}{100\mathrm{mL}}\right)\left(1000\mathrm{\mu g}/\mathrm{mL}\right)$

Concentration of Fe =localid="1663670473584" ${}^{-}100\mathrm{\mu g}/\mathrm{mL}$

localid="1663670938364" $\mathrm{Concentration}\mathrm{of}\mathrm{Fe}=\left(\frac{\mathrm{Volume}\mathrm{of}\mathrm{solution}\mathrm{diluted}}{\mathrm{Volume}\mathrm{of}\mathrm{stock}\mathrm{solution}}\right)(\mathrm{New}\mathrm{concentration}\mathrm{of}\mathrm{Fe})$

$\mathrm{Consider}\mathrm{new}\mathrm{concentration}\mathrm{of}\mathrm{Fe}=100\mathrm{\mu g}/\mathrm{mL}$

$\mathrm{Consider}\mathrm{new}\mathrm{concentration}\mathrm{of}\mathrm{Fe}=\left(\frac{3\mathrm{mL}}{100\mathrm{mL}}\right)\left(100\mathrm{\mu g}/\mathrm{mL}\right)$

$\mathrm{Consider}\mathrm{new}\mathrm{concentration}\mathrm{of}\mathrm{Fe}=3\mathrm{\mu g}/\mathrm{mL}$