Question

Rearrange the following in the order of increasing mass and choose the correct order (Atomic mass, $N=14,O=16,Cu=63$)
(I) 1 molecule of oxygen
(2) 1 atom of nitrogen
(3) $1\times {10}^{-10}gm$ molecule of oxygen
(4) $1\times {10}^{-10}gm$ of copper

• A. $II<I<IV<III$
• B. $IV<III<II<I$
• C. $II<III<I<IV$
• D. $III<IV<I<II$

Answer 1

The correct option is A $II<I<IV<III$

(1) $1$ molecule of oxygen=$O_2$

$\therefore$ Mass of $O_2=\frac{16\times 2}{N_A}=\frac{32g}{N_A}=\frac{32}{6.022\times {10}^{23}}=5.3\times {10}^{-23}g$

(2) $1$ atom of Nitrogen=$\frac{1}{N_A}$ moles

$=\frac{1}{6.022\times {10}^{23}}$ moles

$=1.66\times {10}^{-24}$moles

$\therefore$ Mass of $1$ atom of Nitrogen=$1.66\times {10}^{-24}\times 14$

$=23.2\times {10}^{-24}g$

(3) $1\times {10}^{-10}gm$ molecule of oxygen=$1\times {10}^{-10}$ moles of $O_2$

$\therefore$ Mass of $1\times {10}^{-10}gm$ molecule of oxygen=$1\times {10}^{-10}\times 32$

$=3.2\times {10}^{-9}g$

(4) Mass of copper$=1\times {10}^{-10}gm$

Comparing the masses in (1), (2), (3) and (4)

we get, $\left(2\right)<\left(1\right)<\left(4\right)<\left(3\right)$

Therefore, answer is $\left(II\right)<\left(I\right)<\left(IV\right)<\left(III\right)$.