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Q45) 1.12 mL of gas is produced at S.T.P. by the action of 4.12 mg of alcohol, ROH, with methyl magnesium bromide. The molecular mass of alcohol is .. [ IIT 1993]


ROH + CH3MgBr → CH4 + Mg(OH)Br
Thus One mole of ROH gives 1 mole=22400 mL of CH4
1.12 ml is obtained from 4.12 mg
∴ 22400 mL will be obtained form=82400 mg=82.4 g
Answer: (c)