Question

69 mg of an alcohol $(R-OH)$ was treated with $C{H}_{3}MgBr$ and the gas evolved was measured to be 33.6 ml at STP. The molar mass (g/mol) of alcohol is .

Answer 1

$\underset{1mol}{R-OH}+C{H}_{3}MgBr\to \underset{1mol}{C{H}_4}+Mg\left(Br\right)\left(OR\right)$

1 mole at STP = 22.4 L
Let , x be the molar mass.
$\frac{0.069}{x}=\frac{33.6}{22400}$
$\Rightarrow x=46$