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Q)

$1.12\;mL$ of a gas is produced at STP by the action of $4.12\;mg$ of alcohol ROH with methyl magnesium iodide. The molecular mass of alcohol is

$(a)\;16.0 \\ (b)\;41.2 \\(c)\;82.4 \\(d)\;156.0 $

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A)
$ROH + CH_3 MgBr \longrightarrow CH_4 +BrMgOR$
$22400 mL$ of $CH_4$ produced at $STP=1\;mol \;of\;ROH$
$1.2\;mL$ of $CH_4$ produced at $STP=\large\frac{1.12}{22400}$$= 5 \times 10^{-5}\;mol$
Now molecular mass of $ROH= \large\frac{4.12 \times 10^{-3}}{5 \times 10^{-5}}$
$\therefore $ mol mass of $ROH=82.4$
Hence c is the correct answer.
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