Consider the above sequence of reactions. 151 | vedclass

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(C) The molar mass of $2-$bromopentane $(C_5H_{11}Br)$ is $(5 	imes 12) + (11 	imes 1) + 80 = 60 + 11 + 80 = 151 \ g \ mol^{-1}$.Given mass of $2-$b

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$184$

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(C) The molar mass of $2-$bromopentane $(C_5H_{11}Br)$ is $(5 	imes 12) + (11 	imes 1) + 80 = 60 + 11 + 80 = 151 \ g \ mol^{-1}$.Given mass of $2-$bromopentane = $151 \ g$,so moles of $2-$bromopentane = $1 \ mol$.Reaction $1$: $2-$bromopentane $\xrightarrow{	ext{alc. KOH}}$ pent-$2-$ene $(P)$ + $HBr$.Since the yield of $P$ is $80 \%$,moles of $P$ formed = $0.8 \ mol$.Reaction $2$: Pent-$2-$ene $(P)$ $\xrightarrow{Br_2}$ $2,3-$dibromopentane $(Q)$.Since the yield of $Q$ is $100 \%$,moles of $Q$ formed = $0.8 \ mol$.The molar mass of $Q$ $(C_5H_{10}Br_2)$ is $(5 	imes 12) + (10 	imes 1) + (2 	imes 80) = 60 + 10 + 160 = 230 \ g \ mol^{-1}$.Mass of $Q$ = $	ext{moles} 	imes 	ext{molar mass} = 0.8 \ mol 	imes 230 \ g \ mol^{-1} = 184 \ g$.

Consider the above sequence of reactions. $151 \ g$ of $2-$bromopentane is made to react. The yield of the major product $P$ is $80 \%$ whereas the yield of $Q$ is $100 \%$. The mass of product $Q$ obtained is $........ \ g$. (Given molar mass in $g \ mol^{-1}$: $H: 1, C: 12, Br: 80$)

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