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Question

69 mg of an alcohol (ROH) 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  .


Solution

ROH1 mol+CH3MgBrCH41 mol+Mg(Br)(OR)

1 mole at STP = 22.4 L
Let , x be the molar mass.
0.069x=33.622400
x=46

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