Arrange the following in the order of increasing mass (atomic mass: $\mathrm{O}=16, \mathrm{Cu}=63,…

Arrange the following in the order of increasing mass (atomic mass: $\mathrm{O}=16, \mathrm{Cu}=63, \mathrm{~N}=14$ )
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
  1. $\mathrm{II} < \mathrm{I} < \mathrm{III} < \mathrm{IV}$
  2. $\mathrm{I} < \mathrm{II} < \mathrm{III} < \mathrm{IV}$
  3. $\mathrm{III} < \mathrm{II} < \mathrm{IV} < \mathrm{I}$
  4. $\mathrm{IV} < \mathrm{II} < \mathrm{III} < \mathrm{I}$

Solution

Mass of $6.023 \times 10^{23}$ atoms of oxygen $=16 \mathrm{~g}$ Mass of one atom of oxygen
$=\frac{16}{6.023 \times 10^{23}}=2.66 \times 10^{-23} \mathrm{~g}$
Mass of $6.023 \times 10^{23}$ atoms of nitrogen $=14 \mathrm{~g}$
Mass of one atom of nitrogen
$=\frac{14}{6.023 \times 10^{23}}=2.32 \times 10^{-23} \mathrm{~g}$
Mass of 1 mole of oxygen $=16 \mathrm{~g}$ Mass of $1 \times 10^{-10}$ mole of oxygen $=16 \times 10^{-10}$ Mass of 1 mole of copper $=63 \mathrm{~g}$ 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 $\mathrm{II} < \mathrm{I} < \mathrm{II} \mathrm{I} < \mathrm{IV}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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