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QJEEM23Ch692e: JEE Main 2023 Chemistry - Class 11 JEEM23Ch692e Solution: 17 mg of a hydrocarbon (M.F. ) takes up 8.40 mL of the H gas…

17 mg of a hydrocarbon (M.F. C10H16\mathrm{C_{10}H_{16}} ) takes up 8.40 mL of the H 2_2 gas measured at 0 ∘^\circ C and 760 mm of Hg. Ozonolysis of the same hydrocarbon yields

JEE Main 2023 (Online) 29th January Morning Shift Chemistry - Hydrocarbons Question 49 English

The number of double bond/s present in the hydrocarbon is ___________.

Step-by-Step Verified Solution

Moles of hydrocarbon =17×10−3136=1.25×10−4=\frac{17 \times 10^{-3}}{136}=1.25 \times 10^{-4}

Mole of H2\mathrm{H}_{2} gas

⇒1×8.401000=n×0.0821×273\Rightarrow 1 \times \frac{8.40}{1000}=\mathrm{n} \times 0.0821 \times 273

⇒n=3.75×10−4\Rightarrow \mathrm{n}=3.75 \times 10^{-4}

Hydrogen molecule used for 1 molecule of hydrocarbon is 3

=3.75×10−41.25×10−4=3=\frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}}=3

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