1.12 ml of a gas is produced at STP by the action of 4.12 mg of alcohol, with methyl magnesium iodide. The molecular mass of alcohol is

 $\underset{1\mathrm{mol}}{\mathrm{ROH}}+{\mathrm{CH}}_3\mathrm{MgI}\to \underset{1\mathrm{mol}=22400\mathrm{cc}}{{\mathrm{CH}}_{4\left(\mathrm{g}\right)}+\mathrm{Mg}{⟨}_{\mathrm{I}}^{\mathrm{OR}}}$

1.12 mL of a gas is obtained from 4.12 mg of alcohol

$\therefore$ 22400 mL will be obtained from        ?

                       Molecular weight of  alcohol   =$\frac{4.12}{1.12}\times 22400\mathrm{mg}=824\times {10}^2\mathrm{mg}=82.4\mathrm{g}$