69 mg of an alcohol (R−OH) was treated with CH3MgBr and the gas evolved was measured to be 33.6 ml at STP. The molar mass (g/mol) of alcohol is .
R−OH1 mol+CH3MgBr→CH41 mol+Mg(Br)(OR)
1 mole at STP = 22.4 L
Let , x be the molar mass.