A sample of $4.5 m g$ of an unknown monohydric alcohol, R–OH was added to methylmagnesium iodide. A gas is evolved and is collected and its volume measured to be $3.1 m L .$ The molecular weight of the unknown alcohol is______ $g / m o l .$
[Nearest integer]

33.0

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33.00

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Solution

$R - O H + C H_{3} M g I \Rightarrow R - O M g I + C H_{4}$

moles of alcohol (ROH) = moles of $C H_{4}$
At STP, [Assuming STP]
$1 mole$ corresponds to $22.7 L$

Hence, $3.1 m L = \frac{3.1}{22700} mol$

So, moles of alcohol = $\frac{3.1}{22700} mol$

$\Rightarrow \frac{- 3}{22700} = \frac{4.5 \times 10^{- 3}}{M}$

$M = 33 g/mol$

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