A sample of 4.5 mg 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 mL. The molecular weight of the unknown alcohol is ____ g/mol. [Nearest integer]

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answer is 33.

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Detailed Solution

$ROH + {CH}_{3} MgI \to ROMgI + {CH}_{4} (g)$

moles of CH₄ = moles of ROH

$\begin{array}{l} \Rightarrow \frac{V}{22400} = \frac{m}{M \cdot M} (Assuming NTP Condition) \\ \Rightarrow \frac{3.1}{22400} = \frac{4.5 \times 10^{- 3}}{M \cdot M} \\ \Rightarrow MM = 32.51 \end{array}$

Nearest Integer = 33

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