A $100 mL$ solution of $C H_{3} C H_{2} M g B r$ on treatment with methanol produces $2.24 m L$ of a gas at $S T P$ .
The weight of gas produced is $m g$ . [nearest
integer]

3.00

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(03.00)

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3.0

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Solution

$C H_{3} - C H_{2} - M g B r + C H_{3} O H \to C H_{3} - C H_{3} + M g B r (O C H)_{3}$
As $2.24 m l$ is formed at STP.
Number of moles of ethane gas produced
$= \frac{2.24 \times 10^{- 3} L}{22.4 L}$
$= 10^{- 4}$
Mass of ethane produced $= 10^{- 4} m o l \times 30 g / m o l = 3 \times 10^{- 3} g = 3 m g$

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A 100 mL solution of CH3CH2MgBr on treatment with methanol produces 2.24 mL of a gas at STP. The weight of gas produced is mg. [nearest integer]

CH3 –CH2 –MgBr+CH3OH → CH3 –CH3+MgBr(OCH)3 As 2.24 ml is formed at STP. Number of moles of ethane gas produced =2.24×10−3L22.4 L =10−4 Mass of ethane produced =10−4mol×30 g/mol=3×10−3g=3 mg

A $100 mL$ solution of $C H_{3} C H_{2} M g B r$ on treatment with methanol produces $2.24 m L$ of a gas at $S T P$ .
The weight of gas produced is $m g$ . [nearest
integer]