A) 80
B) 160
C) 240
D) 320
Correct Answer: A
Solution :
\[\underset{\text{42 }gms}{\mathop{\underset{\text{1 mole}}{\mathop{\underset{\text{Propane}}{\mathop{C{{H}_{3}}-CH}}\,}}\,}}\,=\,\,C{{H}_{2}}\,\,+\underset{\underset{42\text{ }gms}{\mathop{1\,\text{mole}}}\,}{\mathop{B{{r}_{2}}}}\,\to C{{H}_{3}}-\underset{\text{1, 2-dibromo propane}}{\mathop{\underset{Br\,\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,-\underset{Br\,\,\,\,}{\mathop{\underset{|\,\,\,\,\,\,\,\,}{\mathop{C{{H}_{2}}}}\,}}\,}}\,\] \[\because \] 42 gms of propene reacts with 160 gms of bromine. \[\therefore \] 21gms of propene \[\frac{160}{42}\times \,21=80\,gms\].
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