375 mg of an alcohol reacts with require...

375 mg of an alcohol reacts with required amount of methyl magnesium bromide and releases 140 mL of methane gas at STP The alcohol is,

ethanol

n-butanol

methanol

n-propanol

Text Solution

Verified by Experts

The correct Answer is:

140 mL of `CH_4` is produced from alcohol = 375mg.
`therefore 22400 mL` of `CH_(4)` will be produced from alcohol `= (375)/1000 xx (22400)/140 = 60 g "mol"^(-1)`
Thus, alcohol with molecular mass = 60 g `mol^(-1)` is `CH_(3)CH_(2)CH_(2)OH` (n-propanol)

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