Among the following, choose the ones with equal number of atoms.

A. 212 g of Na2CO3(s) [molar mass = 106 g]
B. 248 g of Na2O(s) [molar mass = 62 g]
C. 240 g of NaOH(s) [molar mass = 40 g]
D. 12 g of H2(g) [molar mass = 2 g]
E. 220 g of CO2(g) [molar mass = 44 g]

Choose the correct answer from the options given below :

To find the options with an equal number of atoms, we need to calculate the total number of atoms in each given sample. The number of atoms in a sample can be calculated using the formula:

$\text{Number of atoms}=\; \backslash text\{Number\; of\; moles\; of\; compound\}\; \backslash times\; \backslash text\{Number\; of\; atoms\; per\; molecule/formula\; unit\}\; \backslash times\; N\_A$

where $N_A$ is Avogadro's number ($6.022\times {10}^{23}{\text{mol}}^{-1}$).
Since $N_A$ is a constant, we can compare the quantity:
$\text{Total moles of atoms}=\; \backslash text\{Moles\; of\; compound\}\; \backslash times\; \backslash text\{Atoms\; per\; formula\; unit\}$

Let's calculate this value for each option step-by-step:

A. 212 g of ${\text{Na}}_2{\text{CO}}_3$(s) [molar mass = 106 g/mol]
Moles of ${\text{Na}}_2{\text{CO}}_3$ = $\frac{212}{106}=2\text{mol}$.
One formula unit of ${\text{Na}}_2{\text{CO}}_3$ contains 2 Na + 1 C + 3 O = 6 atoms.
Total moles of atoms = $2\times 6=12\text{moles of atoms}$ (or $12\times N_A$ atoms).

B. 248 g of ${\text{Na}}_2\text{O}$(s) [molar mass = 62 g/mol]
Moles of ${\text{Na}}_2\text{O}$ = $\frac{248}{62}=4\text{mol}$.
One formula unit of ${\text{Na}}_2\text{O}$ contains 2 Na + 1 O = 3 atoms.
Total moles of atoms = $4\times 3=12\text{moles of atoms}$ (or $12\times N_A$ atoms).

C. 240 g of $\text{NaOH}$(s) [molar mass = 40 g/mol]
Moles of $\text{NaOH}$ = $\frac{240}{40}=6\text{mol}$.
One formula unit of $\text{NaOH}$ contains 1 Na + 1 O + 1 H = 3 atoms.
Total moles of atoms = $6\times 3=18\text{moles of atoms}$ (or $18\times N_A$ atoms).

D. 12 g of ${\text{H}}_2$(g) [molar mass = 2 g/mol]
Moles of ${\text{H}}_2$ = $\frac{12}{2}=6\text{mol}$.
One molecule of ${\text{H}}_2$ contains 2 atoms.
Total moles of atoms = $6\times 2=12\text{moles of atoms}$ (or $12\times N_A$ atoms).

E. 220 g of ${\text{CO}}_2$(g) [molar mass = 44 g/mol]
Moles of ${\text{CO}}_2$ = $\frac{220}{44}=5\text{mol}$.
One molecule of ${\text{CO}}_2$ contains 1 C + 2 O = 3 atoms.
Total moles of atoms = $5\times 3=15\text{moles of atoms}$ (or $15\times N_A$ atoms).

Comparing the total moles of atoms:
- A has 12 moles of atoms.
- B has 12 moles of atoms.
- C has 18 moles of atoms.
- D has 12 moles of atoms.
- E has 15 moles of atoms.

Thus, options A, B, and D contain an equal number of atoms.