1.

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

Answer»

ethanol
n-butanol
methanol
n-propanol

Solution :`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.


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