17 mg of a hydrocarbon (M.F. C 10 H 16 ) takes up 8 . 40 mL of the H 2 gas measured at 0 ° C and 760 mm of Hg . Ozonolysis of the same hydrocarbon yields CH 3 - C | | O - CH 3 , H - C | | O - H , H -

Question

17 mg of a hydrocarbon (M.F. C 10 H 16 ) takes up 8 . 40 mL of the H 2 gas measured at 0 ° C and 760 mm of Hg . Ozonolysis of the same hydrocarbon yields CH 3 - C | | O - CH 3 , H - C | | O - H , H - C | | O - CH 2 - CH 2 - C | | O - C | | O - H The number of double bond/s present in the hydrocarbon is

TestHub · Accepted Answer

From the given information:
Moles of hydrocarbon $= \frac{17 \times 10^{- 3}}{136} = 1.25 \times 10^{- 4}$ Mole of $H_{2}$ gas

$\Rightarrow 1 \times \frac{8.40}{1000} = n \times 0.0821 \times 273$

$\Rightarrow n = 3.75 \times 10^{- 4}$

Thus, moles of hydrogen used to hydrogenate double bonds:

$= \frac{3.75 \times 10^{- 4}}{1.25 \times 10^{- 4}} = 3$

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