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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]
R−OH+CH3 Mgl ⇒ R−OMgl+CH4\mathrm{R}-\mathrm{OH}+\mathrm{CH}_{3} \mathrm{Mgl} \Rightarrow \mathrm{R}-\mathrm{OMgl}+\mathrm{CH}_{4}\\R−OH+CH3 Mgl ⇒ R−OMgl+CH4 moles of alcohol (ROH) ≡(\mathrm{ROH}) \equiv(ROH) ≡ moles of CH4\mathrm{CH}_{4}CH4
At STP, [Assuming STP] 1 mole corresponds to 22.7 L22.7 \mathrm{~L}\\22.7 L Hence, 3.1 mL ≡ 3.122700 mol3.1 \mathrm{~mL} \equiv \frac{3.1}{22700} \mathrm{~mol}\\3.1 mL ≡ 227003.1 mol So, moles of alcohol =3.122700=\frac{3.1}{22700}\\=227003.1 ⇒ 3.122700=4.5 × 10−3M\Rightarrow \frac{3.1}{22700}=\frac{4.5 \times 10^{-3}}{\mathrm{M}}⇒ 227003.1=M4.5 × 10−3 M ≃ 33 g / mol\\M \simeq 33 \mathrm{~g} / \mathrm{mol}M ≃ 33 g / mol
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