Hydrocarbons Question 6
Question 6 - 29 January - Shift 1
$17 mg$ of a hydrocarbon (M.F. $C _{10} H _{16}$ ) takes up $8.40 mL$ of the $H_2$ gas measured at $0^{\circ} C$ and 760 $mm$ of $Hg$. Ozonolysis of the same hydrocarbon yields
$CH_3-C-CH_3-, H-C-H, H-C-CH_2-CH_2-C-C-H$
The number of double bond/s present in the hydrocarbon is
Show Answer
Answer: (3)
Solution:
Moles of hydrocarbon $=\frac{17 \times 10^{-3}}{136}=1.25 \times 10^{-4}$
Mole of $H_2$ gas
$\Rightarrow 1 \times \frac{8.40}{1000}=n \times 0.0821 \times 273$
$\Rightarrow n=3.75 \times 10^{-4}$
Hydrogen molecule used for 1 molecule of hydrocarbon is 3
$=\frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}}=3$
