A) 22.4 litre
B) 2.016 litre
C) 2.24 litre
D) 20.16 litre
Correct Answer: B
Solution :
[b] \[\underset{10gm}{\mathop{CaC{{O}_{3}}}}\,\to CaO+C{{O}_{2}}\] |
90% pure 9gm \[=\frac{9}{100}\]mole |
\[CaC{{O}_{3}}\equiv C{{O}_{2}}=0.09\]mole |
At NTP Vol.\[C{{O}_{2}}=0.09\times 22.4=2.016L\]. |
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