$0.037g$ of an alcohol, was added to $C{H}_{3}MgBr$ and the gas evolved measured 11.2mL at STP. The molecular mass of  $R-OH$  will be ________(At STP gram molar volume=22.4lit)

 $$\begin{gathered} R-OH+C{H}_{3}MgBr\underset{ }{\overset{ }{\to }}C{H}_4+Mg\left(OR\right)Br \\ \begin{array}{l}1mole\\ \left(or\right)\\ 22,400mLatSTP\end{array} \end{gathered}$$
$11.2mLC{H}_4$   at STP is formed by 0.037g  $R-OH$
$\therefore$  22,400 mL of   $C{H}_4$  at STP will be formed by $=\frac{0.037}{11.2}\times 22400g$  of  $ROH$ $=74g$