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1.12ml of a gas is produced at STP by the action of alcohol ROH with CH3MgI .The molecular mass of alcohol is?
Dear studentROH + CH3MgI ----> R-CH3 + MgI(OH)Here we need to compare moles of ROH and R-CH3 but here question is incomplete since we are not given the weight of ROH taken.
1.12 mL is obtained from 4.12 mg22400 mL will be obtained from
ROH + CH3MgI→CH4 + RO−MgI+Volume of gas produced = 112 mLMass of alcohol used = 412 mg1 mole of alcohol produce 1 mole of CH4.So, 22.4 L CH4 will be produced by the molecular weight of alcohol.112 mL CH4 is produced from 412 mg of alcoholSo, 22400 mL of CH4 will be produced by = 412112×22400=82400 mg of alcohol = 82.4 g of ROHSo,
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