| I. 0.5 mole of 03 |
| II. 0.5 g atom of oxygen |
| III. \[3.011\times {{10}^{23}}\]molecules of\[{{O}_{2}}\] |
| IV. 5.6 L of\[C{{O}_{2}}\]at STP |
A) II<IV<III<I
B) IV<III<I<II
C) II<I<IV<III
D) I<II<III<IV
Correct Answer: A
Solution :
I. 0.5 mole \[{{O}_{3}}=24g{{O}_{3}}\] II. 0.5 g atom of oxygen\[=8\,g\,{{O}_{2}}\] III. \[\frac{3.011\times {{10}^{23}}}{6.022\times {{10}^{23}}}\times 32=16\,g\,{{O}_{2}}\] IV. \[\frac{5.6}{22.4}\times 44\,g\,C{{O}_{2}}=11\,g\,C{{O}_{2}}\] Hence, the order is \[I>III>IV>II\]
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