1.12 $\mathrm{mL}$ of a gas is produced at S.T.P. by the action of $4.12 \mathrm{mg}$ of alcohol…

1.12 $\mathrm{mL}$ of a gas is produced at S.T.P. by the action of $4.12 \mathrm{mg}$ of alcohol $\mathrm{ROH}$ with methyl magnesium Iodide. The molecular mass of alcohol is
  1. $16.0$
  2. $41.2$
  3. $82.4$
  4. $156.0$

Solution

Let the alcohol be
$\mathrm{ROH}$ and $\mathrm{x}$ its molecular weight
$\mathrm{ROH}+\mathrm{CH}_{3} \mathrm{MgI} ightarrow \mathrm{CH}_{4}+\mathrm{ROMgI}$
$\mathrm{xg}$ $\quad\quad\quad\quad\quad\quad\quad\quad\mathrm{16g}$
$\frac{4.12}{1000} \mathrm{~g}$ of alcohol will produce $\frac{16}{x} \times \frac{4.12}{1000} g$ of methane
Methane actually obtained is $=\frac{16 \times 1.12}{22400} \mathrm{~g}$
$\therefore \frac{16 \times 4.12}{x \times 1000}=\frac{16 \times 1.12}{22400}$
$=x=82.4$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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