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.
Comments
Leave a comment