3750 mg of an alcohool reacts with required amount of methyl magnesium bromide and release 140mL of methane gas at STP. The alcohol is `:`

Options

A. ethanol
B. n-butanol
C. methanol
D. n-propanol

Answer

D

Explanation

`underset("1 mol")(ROH)+CH_(3)MgBrrarr underset("= 22400 mL at STP ")underset("1 mol ")(CH_(4)+Mg(OR)Br)`
140 mL of `CH_(4)` is produced from alchohol
= 375 mg = 0.375 g
`therefore" 22400 mL of CH"_(4)" will be produced from"`
`"alcohol"=(0.375)/(140)xx22400=60g`
`therefore" Molar mass of alcohol = 60 g mol"^(-1)`
`M(C_(2)H_(5)OH)=46, M(C_(4)H_(9)OH)=74,`
`M(CH_(3)OH)=32, M(CH_(3)CH_(2)CH_(2)OH)=60`
Hence, the alcohol is n-propanol.