81.4 gm/mole
- Swarna Kamal Dhyawala sorry 82.4gm/moleUpvote·0· Reply ·2013-02-28 07:45:26
1.12 ml of a gas is produced at S.T.P. by the action of 4.12 mg of alcohol, ROH, with methyl magnesium iodide. find the molecular mass of alcohol.
81.4 gm/mole
ROH+CH3MgI= CH4+ROMg+I
MOLES OF CH4(GAS)≡5*10-5
now (5*10-5) mole of ROH is 4.12 mg
1 mole of ROH = 82.4 gm