Q30 / 90 · JEE Main 2012 (07 May Online) · Chemistry
The ratio of number of oxygen atoms $(\mathrm{O})$ in $16.0 \mathrm{~g}$ ozone $\left(\mathrm{O}_3\right), 28.0 \mathrm{~g}$ carbon monoxide $(\mathrm{CO})$ and $16.0$ oxygen $\left(\mathrm{O}_2\right)$ is (Atomic mass : $\mathrm{C}=12, \mathrm{O}=16$ and Avogadro's constant $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$ )
- A$3: 1: 2$
- B$1: 1: 2$
- C$3: 1: 1$
- Correct answerD$1: 1: 1$
Solution
$16.0 \mathrm{~g} \mathrm{O}_3=\frac{16}{48}$ mole
$$
\begin{aligned}
& =\frac{16}{48} \times 6.023 \times 10^{23} \text { molecules } \\
& =3 \times \frac{16}{48} \times 6.023 \times 10^{23} \text { atoms } \\
& =6.023 \times 10^{23} \text { atoms } \\
& 28.0 \mathrm{~g} \mathrm{CO}=\frac{28}{28} \text { mole }=1 \text { mole } \\
& =1 \times 6.023 \times 10^{23} \text { molecules } \\
& =1 \times 6.023 \times 10^{23} \text { atoms } \\
& =6.023 \times 10^{23} \text { atoms } \\
& 16.0 \mathrm{~g} \mathrm{O}_2=\frac{16}{16} \text { mole }=1 \mathrm{~mole}
\end{aligned}
$$
$=1 \times 6.023 \times 10^{23}$ molecules
$=1 \times 6.023 \times 10^{23}$ atoms
$=6.023 \times 10^{23}$ atoms
Therefore, the ratio is
$6.023 \times 10^{23}: 6.023 \times 10^{23}: 6.023 \times 10^{23}$
i.e. $1: 1: 1$
Hence, the correct answer is option: 'D'.