0.037g of an alcohol, R-OH, was added to C2H5Mgl and the gas evolved measured 11.2 cc at STP. The molecular mass of the alcohol is

R−OH+C2H5Mgl→C2H6(g)+Mg(OR)I11.2 cc of gas at STP = (11.2/22400)mol = 0.0005 mol. From the equation, 0.0005 mol of gas will be produced from 0.0005 mol of ROH.The mass of 0.0005 mol of ROH = 0.037 g.∴The molar mass of ROH = 0.037g/0.0005 mol = 74 g/mol.