JEE MAIN - Chemistry (2022 - 25th July Evening Shift - No. 20)
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]
Answer
33
Explanation
$$\mathrm{R}-\mathrm{OH}+\mathrm{CH}_{3} \mathrm{Mgl} \Rightarrow \mathrm{R}-\mathrm{OMgl}+\mathrm{CH}_{4}$$
moles of alcohol $(\mathrm{ROH}) \equiv$ moles of $\mathrm{CH}_{4}$
At STP, [Assuming STP]
1 mole corresponds to $$22.7 \mathrm{~L}$$
Hence, $$3.1 \mathrm{~mL} \equiv \frac{3.1}{22700} \mathrm{~mol}$$
So, moles of alcohol $$=\frac{3.1}{22700}$$
$$\Rightarrow \frac{3.1}{22700}=\frac{4.5 \times 10^{-3}}{\mathrm{M}}$$
$$M \simeq 33 \mathrm{~g} / \mathrm{mol}$$
moles of alcohol $(\mathrm{ROH}) \equiv$ moles of $\mathrm{CH}_{4}$
At STP, [Assuming STP]
1 mole corresponds to $$22.7 \mathrm{~L}$$
Hence, $$3.1 \mathrm{~mL} \equiv \frac{3.1}{22700} \mathrm{~mol}$$
So, moles of alcohol $$=\frac{3.1}{22700}$$
$$\Rightarrow \frac{3.1}{22700}=\frac{4.5 \times 10^{-3}}{\mathrm{M}}$$
$$M \simeq 33 \mathrm{~g} / \mathrm{mol}$$
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