Question

An experiment requires a solution that is 80% methyl alcohol by volume. What volume of methyl alcohol should be added to 200 mL of water to make this solution?

Answer 1

The percentage of CH₃OH by volume is defined as:

$\phi_{MeOH}=\frac{V_{MeOH}}{V_{MeOH}+V_{H_2O}}*100\%;$

We have to modify this equation:

$V_{MeOH}*\phi_{MeOH}+\phi_{MeOH}*V_{H_2O}=100\%*V_{MeOH};$

$V_{MeOH}(100\%-\phi_{MeOH})=\phi_{MeOH}*V_{H_2O};$

$V_{MeOH}=\frac{\phi_{MeOH}*V_{H_2O}}{100\%-\phi_{MeOH}}=\frac{80\%*200mL}{100\%-80\%}=800mL.$

Thus, 800 mL of pure methyl alcohol should be added to 200 mL of water.