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Q.
An aqueous solution contains $25\%$ ethanol and $50\%$ acetic acid by mass. Calculate the mole fraction of acetic acid in this solution:
J & K CETJ & K CET 2001
Solution:
Mole of $C_{2} H _{5} OH =\frac{25}{46}=0.54$
Mole of $CH _{3} COOH =\frac{50}{60}=0.83$
$ \%$ of $H _{2} O =100-(50+25)=25 $
$\therefore $ Mole of $H _{2} O =\frac{25}{18}=1.38$
Total number of moles $=0.54+0.83+1.39$
$=2.76$
$ \therefore $ Mole fraction of acetic acid
$=\frac{0.83}{2.76}=0.301$