Among the following, choose the ones with equal number of atoms:A. $212\text{g}$ of ${\text{Na}}_2{\text{CO}}_3$ ($\text{s}$) [molar mass = $106\text{g}$ ]B. $248\text{g}$ of ${\text{Na}}_2\text{O}$ ($\text{s}$) [molar mass = $62\text{g}$ ]C. $240\text{g}$ of $\text{NaOH}$ ($\text{s}$) [molar mass = $40\text{g}$ ]D. $12\text{g}$ of ${\text{H}}_2$ ($\text{g}$) [molar mass = $2\text{g}$ ]E. $220\text{g}$ of ${\text{CO}}_2$ ($\text{g}$) [molar mass = $44\text{g}$ ]Choose the correct answer from the options given below:

To find the total number of atoms, we calculate moles of the substance and multiply by the number of atoms in one molecule and Avogadro's number ($N_A$).A: $\text{Moles}=\frac{212}{106}=2$. Atoms per molecule = $6$. Total atoms = $2\times 6\times N_A=12N_A$.B: $\text{Moles}=\frac{248}{62}=4$. Atoms per molecule = $3$. Total atoms = $4\times 3\times N_A=12N_A$.C: $\text{Moles}=\frac{240}{40}=6$. Atoms per molecule = $2$. Total atoms = $6\times 2\times N_A=12N_A$.D: $\text{Moles}=\frac{12}{2}=6$. Atoms per molecule = $2$. Total atoms = $6\times 2\times N_A=12N_A$.E: $\text{Moles}=\frac{220}{44}=5$. Atoms per molecule = $3$. Total atoms = $5\times 3\times N_A=15N_A$.Species B, C, and D all have $12N_A$ atoms.