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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,

A

ethanol

B

n-butanol

C

methanol

D

n-propanol

Text Solution

Verified by Experts

The correct Answer is:
D
The required reaction is

Moles of `CH_(4)(n)` at STP
`= (140)/(22400) = 0.00625` moles
Here, `CH_(3) -` of `CH_(3)MgBr` combines with H-atom of R-OG to give `CH_(4)`.
`therefore n(CH_(4)) = n(R-OH)`
Also, `n = (W)/("M(molar mass)")`
`= 0.00625 = (375 xx 10^(-3))/(M)`
`therefore M = (375)/(0.00625 xx 1000) = 60`
Also, general formula for alcohol
`= C_(n)H_(2n + 1).OH`
Thus, `12 + 2n +1 + 17 = 60 " "( because OH = 16 + 1 = 17)`
`therefore 14n = 60 - 18 = 42`
`therefore n = 3`
Hence, alcohol is n-propanol `(C_(3)H_(7)OH)`.
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