A) 16.0
B) 41.2
C) 82.4
D) 156.0
Correct Answer: C
Solution :
\[\underset{1\,\text{mol}\text{.}}{\mathop{\underset{{}}{\mathop{ROH}}\,}}\,+C{{H}_{3}}MgI\to \underset{1\text{ mol}=22400\,cc}{\mathop{C{{H}_{4}}+Mg\ \ \ \ \begin{matrix} OR \\ I\,\,\, \\ \end{matrix}}}\,\] 1.12 mL is obtained from 4.12 mg \ 22400 mL will be obtained from \[\frac{4.12}{1.12}\times 22400\,mg=84.2\,g\]
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