17 mg of a hydrocarbon (M.F. C10H16) takes up 8.40 mL of the H2 gas measured at 0ºC and 760 mm of Hg. Ozonolysis of the same hydrocarbon yields O CH3–C–CH3 , O H–C–H , O H–C–CH2–CH2–CH2–C–C–H O O The

Question

17 mg of a hydrocarbon (M.F. C10H16) takes up 8.40 mL of the H2 gas measured at 0ºC and 760 mm of Hg. Ozonolysis of the same hydrocarbon yields O CH3–C–CH3 , O H–C–H , O H–C–CH2–CH2–CH2–C–C–H O O The number of double bond/s present in the hydrocarbon is ________.

Prashant Kotian · Accepted Answer

Correct answer: .. C10H16 has DU = 3 Now milli moles of hydrocarbon and H2 gas are as follows respectively: 17 = milli moles of hydrocarbon = .125 10 40  . = .375 mm of H2 gas Hence, they are in ratio 1 : 3 to confirm 3Bonds in molecules

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