A) 16.0
B) 41.2
C) 82.4
D) 156.0
Correct Answer: C
Solution :
| [c] Let the alcohol be ROH and x its molecular weight |
| \[\underset{xg}{\mathop{ROH}}\,+C{{H}_{3}}MgI\to \underset{16g}{\mathop{C{{H}_{4}}}}\,+ROMgI\] |
| \[\frac{4.12}{1000}g\] of alcohol will produce \[\frac{16}{x}\times \frac{4.12}{1000}g\]of methane |
| Methane actually obtained is = \[\frac{16\times 1.12}{22400}g\] |
| \[\therefore \frac{16\times 4.12}{x\times 1000}=\frac{16\times 1.12}{22400}=x=82.4\] |
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