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

ROH1  mol  +  CH3 MgI  →  CH4(g)+ Mg⟨IOR1  mol  =  22400   cc1.12 mL of a gas is obtained from 4.12 mg of alcohol∴ 22400 mL will be obtained from        ?                       Molecular weight of  alcohol   =4.121.12  ×   22400  mg  =824×102mg=  82.4  g