A sample of 4.5mg 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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7 months ago

(Calculation is done considering STP condition)

$R O H + C H_{3} M g l \to C H_{4} + R O M g l$

No. of moles ROH = no. of moles of CH4

$\frac{4.5 * 1 0^{- 3}}{M} = \frac{31}{22400} \Rightarrow 32.52 \approx 33 g m / m o l e$

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