A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each component.

Answer

`x_(H_(2)O)=0.503, x_(C_(2)H_(5)OH)=0.196, x_(CH_(3)COOH)=0.301`

Explanation

Weight of water = 25 g, Moles of water `=(25)/(18)=1.388`
Weight of ethanol = 25 g
Moles of ethanol `(C_(2)H_(5)OH)=(25)/(46)=0.543`
Weight of acetic acid `(CH_(3)COOH) = 50g`
Moles of acetic acid `= (50)/(60)=0.833`
Total number of moles `= 1.388 +0.543 +0.833 =2.764`
Mole fraction of water `=(1.388)/(2.764)=0.503`
Mole fraction of ethanol `=(0.543)/(0.764)=0.196`
Mole fraction of acetic acid `=(0.833)/(2.764)=0.301`