Question

Which of the following contains maximum no. of 0-atoms?

• A. 1 g 0
• B. 1 g $O_2$
• C. 1 g $O_3$
• D. All have same number of atoms.

Answer 1

Answer: (D)

All have same number of atoms.

Answer 2

To determine which option contains the maximum number of oxygen atoms, we calculate the number of oxygen atoms in 1 gram of each substance. Let $N_A$ be Avogadro's number ($6.022 \times 10^{23}$ atoms/mol). The atomic mass of oxygen (O) is approximately 16 g/mol.

1. 1 g O (atomic oxygen):

   • Molar mass of O = 16 g/mol.
   • Number of moles of O = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{1 \text{ g}}{16 \text{ g/mol}} = \frac{1}{16}$ mol.
   • Number of oxygen atoms = Number of moles $\times N_A = \frac{1}{16} N_A$ atoms.

2. 1 g $O_2$ (molecular oxygen):

   • Molar mass of $O_2$ = $2 \times 16 \text{ g/mol} = 32 \text{ g/mol}$.
   • Number of moles of $O_2$ = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{1 \text{ g}}{32 \text{ g/mol}} = \frac{1}{32}$ mol.
   • Each $O_2$ molecule contains 2 oxygen atoms.
   • Number of oxygen atoms = Number of moles of $O_2 \times N_A \times 2 = \frac{1}{32} \times N_A \times 2 = \frac{2}{32} N_A = \frac{1}{16} N_A$ atoms.

3. 1 g $O_3$ (ozone):

   • Molar mass of $O_3$ = $3 \times 16 \text{ g/mol} = 48 \text{ g/mol}$.
   • Number of moles of $O_3$ = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{1 \text{ g}}{48 \text{ g/mol}} = \frac{1}{48}$ mol.
   • Each $O_3$ molecule contains 3 oxygen atoms.
   • Number of oxygen atoms = Number of moles of $O_3 \times N_A \times 3 = \frac{1}{48} \times N_A \times 3 = \frac{3}{48} N_A = \frac{1}{16} N_A$ atoms.

Comparing the number of oxygen atoms in each case:

A) 1 g O: $\frac{1}{16} N_A$ atoms B) 1 g $O_2$: $\frac{1}{16} N_A$ atoms C) 1 g $O_3$: $\frac{1}{16} N_A$ atoms

All three options contain the same number of oxygen atoms.