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» Partial pressure of CO is 0.6 atm in the evolved gas If just sufficient `O_(2)`si introduced in the CONTAINER to combust CO and `H_(2)`completely, then final volume FO gases would be 0.68 L at 1 atm and 273 K 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 If after combustion the mixture of gases at 273 K is passes through KOH(aq) , contraction of 1.344 L would take place Answer :A::B::D