17 , mg of a hydrocarbon (M.F. C_{10}H_{16}) | vedclass

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(A) $1$. Calculate the moles of the hydrocarbon: Molar mass of $C_{10}H_{16} = (10 	imes 12) + (16 	imes 1) = 136 \, g/mol$. Moles of hydrocarbon $=

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$3$

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(A) $1$. Calculate the moles of the hydrocarbon: Molar mass of $C_{10}H_{16} = (10 	imes 12) + (16 	imes 1) = 136 \, g/mol$. Moles of hydrocarbon $= \frac{17 	imes 10^{-3} \, g}{136 \, g/mol} = 1.25 	imes 10^{-4} \, mol$.$2$. Calculate the moles of $H_2$ gas: At $STP$ ($0^{\circ}C$,$760 \, mm \, Hg$),$22400 \, mL$ corresponds to $1 \, mol$. Moles of $H_2 = \frac{8.40 \, mL}{22400 \, mL/mol} = 3.75 	imes 10^{-4} \, mol$.$3$. Determine the number of double bonds: Each double bond consumes $1 \, mol$ of $H_2$ for hydrogenation. Number of double bonds $= \frac{	ext{moles of } H_2}{	ext{moles of hydrocarbon}} = \frac{3.75 	imes 10^{-4}}{1.25 	imes 10^{-4}} = 3$.

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