0.01 mol TNT was completely decomposed as;
$\underset{TNT}{C_{7} H_{5} N_{3} O_{6} (g)} ⟶ CO (g) + H_{2} (g) + N_{2} (g) + C (g)$
The gases evolved occupied 2.24L at constant correct option on the basis of above information.

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If just sufficient $O_{2}$ is introduced in the container to combust CO and $H_{2}$ completely, then final volume of gases would be 0.896 Lat 1 atm and 273 K

Partial pressure of CO is 0.6 atm in the evolved gas

If just sufficient $O_{2}$ is introduced in the container to combust CO and $H_{2}$ completely,then final volume of gases would be 1.68 Lat 1 atm and 273 K

If after combustion the mixture of gases at 273 K is passed through $KOH$ (aq), contraction of 1.344 L would take place

answer is A, B, D.

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Detailed Solution

$\begin{aligned} C_{7} H_{5} N_{3} O_{6} \to 6 CO + \frac{5}{2} H_{2} + \frac{3}{2} N_{2} + C (s) \\ 0.01 \underset{0.1 mole = 2.24 L \Rightarrow P = 1 atm}{\underset{⏟}{0.06 0.025 0.015}} \end{aligned}$