| Arrange the following in the order of increasing mass (atomic mass:\[0 = 16, Cu= 63, N= 14\]) |
| 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 :
| (I) mass of \[6.023\times {{10}^{23}}\]atoms of oxygen \[=16g\] |
| Mass of one atom of oxygen \[=\frac{16}{6.023\times {{10}^{23}}}=2.66\times {{10}^{-23}}g\] |
| (II) Mass of \[6.023\times {{10}^{23}}\]atom of nitrogen \[=14\,g\] |
| Mass of one atom of nitrogen\[=\frac{14}{6.023\times {{10}^{23}}}=2.32\times {{10}^{-23}}g\] |
| (III) Mass of 1 mole of oxygen =16g |
| Mass of \[1\times {{10}^{-10}}\] mole of oxygen \[=16\times {{10}^{-10}}\] |
| (IV) Mass of 1 mole of copper=63g |
| 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 |
| \[\operatorname{II}<I<III<IV.\] |
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