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 $V_1$.