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A 100 \mathrm{~mL} solution of \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{MgBr} on treatment with methanol produces 2.24 \mathrm{~mL} of a gas at STP. The weight of gas produced is ___________ mg. [nearest integer]

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

The reaction will be -

\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{MgBr}+\mathrm{CH}_3 \mathrm{OH} \longrightarrow \mathrm{CH}_3-\mathrm{CH}_3+\mathrm{Mg}(\mathrm{OH}) \mathrm{Br}

                                          Methanol

Molar mass of \mathrm{CH}_3 \mathrm{CH}_3=12 \times 2+1 \times 6=30 \mathrm{~g} / \mathrm{mol}

Now, 2.24 \mathrm{ml}=2.24 \times 10^{-3} \mathrm{~L}=22.4 \times 10^{-4} \mathrm{~L}

We know, 1 \mathrm{~mol}=22.4 \mathrm{~L}

So, \mathrm{22.4 \times 10^{-4} L=1 \times 10^{-4} \mathrm{~mol}}

Then, \text { mole }=\frac{\text { mass }}{\text { Molar mass }}

\Rightarrow \mathrm{Mass=Mole\times Molar \; mass}

\Rightarrow \mathrm{Mass=1\times10^{-4}\times 30\; g}

\Rightarrow \mathrm{Mass=3\times10^{-3}\; g}

\Rightarrow \mathrm{Mass=3\; mg}

Answer = 3

Posted by

Sayak

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