| Arrange the following in the order of increasing mass (atomic mass:\[O=16,\text{ }Cu=63,\text{ }N=14\]) |
| I. one atom of oxygen |
| II. one atom of nitrogen |
| III. \[1\times {{10}^{-10}}\] mole of oxygen |
| IV. \[1\times {{10}^{-10}}\] mole of copper |
A) II<I<III<IV
B) I<II<III<IV
C) III<II<IV<I
D) IV<II<III<I
Correct Answer: A
Solution :
[a]| Mass of\[~6.023\times {{10}^{23}}\] atoms of oxygen = 16 g Mass of one atom of oxygen \[=\frac{16}{6.023\times {{10}^{23}}}=2.66\times {{10}^{-23}}g\] |
| Mass of \[6.023\times {{10}^{23}}\] atoms of nitrogen =14g |
| Mass of one atom of nitrogen \[=\frac{14}{6.023\times {{10}^{23}}}=2.32\times {{10}^{-23}}g\] |
| Mass of 1 mole of oxygen = 16 g |
| Mass of \[1\times {{10}^{-10}}\] mole of oxygen \[=16\times {{10}^{-10}}\] |
| Mass of 1 mole of copper = 63 g |
| Mass of \[1\times {{10}^{-10}}\]mole of copper \[=63\times 1\times {{10}^{-10}}\]\[=63\times {{10}^{-10}}\] |
| So, the order of increasing mass is II < I < III < IV. |
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