| Arrange the following (I to IV) in the order of increasing masses. |
| I. 0.5 mole of \[{{O}_{3}}\] |
| II. \[3.01\times {{10}^{23}}\] molecules of \[{{O}_{2}}\] |
| III. 5.6 L of \[C{{O}_{2}}\] at STP |
| IV. 0.5 g atom of \[{{O}_{2}}\] |
A) IV < III < II < I
B) II < I < III < IV
C) II = I < III < IV
D) I < III < IV < II
Correct Answer: A
Solution :
| [a] Idea: To solve this problem student is ad vised to keep in mind the clear understanding of mole concept. |
| \[\text{Number of moles}=\frac{\text{Wt}.\text{ in gram}}{\text{Molecular mass}}\] |
| (I) 0.5 mol of \[{{O}_{3}}=24g{{O}_{3}}\] |
| (II) \[3.01\times {{10}^{23}}\] molecules of \[{{O}_{2}}=16g{{O}_{2}}\] |
| (III) 5.6L of \[C{{O}_{2}}=\frac{5.6}{22.4}\times 44=11g\,\,C{{O}_{2}}\] |
| (IV) 0.5 g atom of \[{{O}_{2}}=8g\] |
| Hence, IV < III < II < I |
| TEST Edge: These types of problems are asked generally in JEE Main to judge the basic mole concept so students are advised to go through clear study of equivalent wt., stoichiometry etc. Question involving this concept may also be asked. |
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