Answer to Question #99137 in General Chemistry for Amy

Question #99137
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?
1
Expert's answer
2019-11-21T06:26:06-0500

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.


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