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0.01 mol of a gaseous compound $C_{2} H_{2} O_{x}$ was treated with 224 mL of $O_{2}$ at NTP. After complete combustion the total volume of gases is 560 mL at NTP. On treatment with $K O H$ solution the volume decreases to 112 mL. The value of x is:

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answer is 4.

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Detailed Solution

Volume of $C O_{2} +$ volume of remaining oxygen=560 mL Or Volume of remaining oxygen $= 112 m L$

$C_{2} H_{2} O_{x} (g) + (\frac{5 - x}{2}) O_{2} (g) \overset{Δ}{\to} 2 C O_{2} (g) + H_{2} O (l)$
224 mL	224 mL	0	-
0	112 mL	448 mL	-

For used moles of $O_{2} : (\frac{5 - x}{2}) \times 0.01 = \frac{0.01}{2}$ or $x = 4$

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