Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction: C_(6)H_(6)(g)+3H_(2(g)) rarr C_(6)H_(12(g)) A mixture of C_(6)H_(6) and excess H_(2) has a pressure of 60 mm of Hg in an unknown volume. After the gas has been passed over a nickel catalyst and all the benzene converted to cyclohexane. the pressure of the gas was 30 mm of Hg in the same volume and temperature. The fraction of C_(6)H_(6) (by volume) present in the original volume is:

Gaseous benzene reacts with hydrogen gas in presence of a nickel catal

1.	Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction: C_(6)H_(6)(g)+3H_(2(g)) rarr C_(6)H_(12(g)) A mixture of C_(6)H_(6) and excess H_(2) has a pressure of 60 mm of Hg in an unknown volume. After the gas has been passed over a nickel catalyst and all the benzene converted to cyclohexane. the pressure of the gas was 30 mm of Hg in the same volume and temperature. The fraction of C_(6)H_(6) (by volume) present in the original volume is:
Answer» `1//3` `1//4` `1//5` `1//6` Solution :Suppose INITIALLY, pressure of `C_(6)H_(6)(G) =P_(1) mm and" that of "H_(2)(g) =P_(2) mm` `THEREFORE P_(1)+P_(2)=60 mm"….(1)"` After the reaction Pressure of `C_(6)H_(6)(g)=0` Pressure of `H_(2)(g)=P_(2)-3P_(1)` Pressure of `C_(6)H_(12)(g)=P_(1)` Total pressure `=P_(2)-3P_(1)+P_(1)=30 mm` Or, `P_(2)-2P_(1)=30 mm"....(2)"` Solving (1) & (2) we GET `P_(1)=10 mm` `P_(2)=50 mm` Fraction of `C_(6)H_(6)(g)` volume = Fraction of moles = Fraction of pressure `=(10)/(60)=(1)/(6)` Hence, (D) is the CORRECT answer.