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Molality

50

Question

50% of the reagent is used for dehydrohalogenation of 6.45 g $C H_{3} C H_{2} C l$ . What will be the weight of the main product obtained?
[Atomic mass of H, C, and Cl are 1, 12 and $35.5 g m o l^{- 1}$ respectively]

A

1.4 g

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B

0.7 g

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C

2.8 g

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D

5.6 g

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Solution

The correct option is D 1.4 g
Dehydrohalogenation of $C H_{3} C H_{2} C l$ gives ethene ( $C_{2} H_{4}$ ) as the main product as:

$C H_{3} C H_{2} C l + a l c . K O H \to C_{2} H_{2} + K C l + H_{2} O$

Molar mass of $C H_{3} C H_{2} C l = 64.5 g / m o l$ and Molar mass of ethene ( $C_{2} H_{4}$ ) = $28 g / m o l$

Number of moles of $C H_{3} C H_{2} C l = \frac{6.45}{64.5} = 0.1 m o l$

Since $1 m o l$ of $C H_{3} C H_{2} C l$ will give $1 m o l$ of ethene. However, only 50% of the reagent is used. So, only 50% of product will be formed.

$∴$ 0.05 mol of $C H_{3} C H_{2} C l$ will react to give 0.05 mol of ethene.

mass of ethene in $0.05 m o l$ =moles x molecular wt.= $0.05 m o l \times 28 g / m o l = 1.4 g$

Option A is correct.

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