A) 0.196
B) 0.301
C) 0.392
D) 0.503
Correct Answer: B
Solution :
Mole of \[{{C}_{2}}{{H}_{5}}OH=\frac{25}{46}=0.54\] Mole of\[C{{H}_{3}}COOH=\frac{50}{60}=0.83\] % 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 \[=\frac{0.83}{2.76}=0.301\]You need to login to perform this action.
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