JEE MAIN - Chemistry (2023 - 29th January Morning Shift - No. 21)
17 mg of a hydrocarbon (M.F. $$\mathrm{C_{10}H_{16}}$$) 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
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The number of double bond/s present in the hydrocarbon is ___________.
Answer
3
Explanation
Moles of hydrocarbon $=\frac{17 \times 10^{-3}}{136}=1.25 \times 10^{-4}$
Mole of $\mathrm{H}_{2}$ gas
$\Rightarrow 1 \times \frac{8.40}{1000}=\mathrm{n} \times 0.0821 \times 273$
$\Rightarrow \mathrm{n}=3.75 \times 10^{-4}$
Hydrogen molecule used for 1 molecule of hydrocarbon is 3
$=\frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}}=3$
Mole of $\mathrm{H}_{2}$ gas
$\Rightarrow 1 \times \frac{8.40}{1000}=\mathrm{n} \times 0.0821 \times 273$
$\Rightarrow \mathrm{n}=3.75 \times 10^{-4}$
Hydrogen molecule used for 1 molecule of hydrocarbon is 3
$=\frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}}=3$
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