# How to dilute a stock solution when the resulting solution has a smaller volume? [closed]

Describe how you would dilute $2.0~\mathrm{L}$ of a $0.85~\mathrm{mol/L}$ magnesium hydroxide stock solution to make $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ magnesium hydroxide.

I know the $$c_1 V_1 = c_2 V_2$$ formula and that magnesium hydroxide's molar mass is $58.32~\mathrm{g/mol}$. However, I am confused because $V_2$ is lower than $V_1$.

## closed as off-topic by Todd Minehardt, ron, Martin - マーチン♦Jan 22 '16 at 3:56

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Approach this problem as if you were in a lab and you actually had a bottle of $2.0~\mathrm{L}$ of a $0.85~\mathrm{mol/L}$ magnesium hydroxide stock solution, and you wanted to make a new solution of $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ magnesium hydroxide.
You do not need to dilute the entire stock, but just a portion of the stock solution. Describe what volume of stock solution you would take out, and how you would dilute it, using the $$c_1 V_1 = c_2 V_2$$ formula, to reach $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ of magnesium hydroxide.