100 ml of 0.1 M acetic acid is completely neutralised using a standard solution of $N a O H$ . The volume of ethane obtained at STP after the complete electrolysis of the resulting solution is :

112 ml

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56 ml

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224 ml

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560 ml

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Solution

The correct option is D 112 ml
$C H_{3} C O O H + N a O H \to C H_{3} C O O N a + H_{2} O$
Number of moles of sodium acetate
$= \frac{100 m l}{1000 m l / L} \times 0.1 m o l / L = 0.01 m o l$
$2 C H_{3} C O O N a h y d r o l y s i s - ------ \to e l e c t r o l y s i s C H_{3} - C H_{3} + 2 C O_{2} + 2 N a O H + H_{2} ↑$

Number of moles of ethane are one half the number of moles of sodium acetate. Number of moles of ethane $= \frac{0.01}{2} = 0.005$ moles.
At STP, 1 mole of ethane occupies a volume of 22400 ml. Hence, 0.005 moles of ethane will occupy a volume of $0.005 \times 22400 = 112$ ml.

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