A) 0.1 mole of \[C{{O}_{2}}\] gas
B) 11.2L of \[C{{O}_{2}}\]gas at STP
C) 22 got \[C{{O}_{2}}\]gas
D) \[22.4\times {{10}^{3}}mL\]of\[C{{O}_{2}}\]gas at STP
Correct Answer: A
Solution :
0.1 mole of \[C{{O}_{2}}\] \[\frac{11.2}{22.4}=0.5\]mole of \[C{{O}_{2}}\] \[\frac{22}{44}=0.5\]mole of \[C{{O}_{2}}\] \[\frac{22.4\times {{10}^{3}}}{22400}=1\]mole of \[C{{O}_{2}}\] Hence, the number of moles of \[C{{O}_{2}}\]is least ie, 0.1
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