0.01 mol TNT was completely decomposed as, {:(C_(7)H_(5)N_(3)O_(6)(s), rarr, CO(g),+,H_(2)(g),+, N_(2)(g),+,C(s)),(""TNT,,,,,,,,):} The gases evolved occupied 2.24 L at constantpressure and 273 K in the eudiometer . Select the correct option on the basis of above information.

0.01 mol TNT was completely decomposed as, {:(C_(7)H_(5)N_(3)O_(6)(s),

1.	0.01 mol TNT was completely decomposed as, {:(C_(7)H_(5)N_(3)O_(6)(s), rarr, CO(g),+,H_(2)(g),+, N_(2)(g),+,C(s)),(""TNT,,,,,,,,):} The gases evolved occupied 2.24 L at constantpressure and 273 K in the eudiometer . Select the correct option on the basis of above information.
Answer» <html><body><p>Partial pressure of CO is 0.6 atm in the evolved gas<br/>If just sufficient `O_(2)`si introduced in the <a href="https://interviewquestions.tuteehub.com/tag/container-20566" style="font-weight:bold;" target="_blank" title="Click to know more about CONTAINER">CONTAINER</a> to combust CO and `H_(2)`completely, then final volume <a href="https://interviewquestions.tuteehub.com/tag/fo-457881" style="font-weight:bold;" target="_blank" title="Click to know more about FO">FO</a> gases would be 0.68 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a> at 1 atm and <a href="https://interviewquestions.tuteehub.com/tag/273-1832932" style="font-weight:bold;" target="_blank" title="Click to know more about 273">273</a> K<br/>If just sufficient `O_(2)` is introduced in the container to combust CO and `H_(2)`completely, then fianl volumeof gases would be 0.896 L at 1 atm and 273 K<br/>If after combustion the mixture of gases at 273 K is passes through KOH(aq) , contraction of 1.344 L would take place</p>Answer :A::B::D</body></html>