Among the following, choose the ones with equal number of atoms: A. $212\text{ g of } \text{Na}_2\text{CO}_3(s)$ [molar mass = $106\text{ g}$] B. $248\text{ g of } \text{Na}_2\text{O}(s)$ [molar mass = $62\text{ g}$] C. $240\text{ g of } \text{NaOH}(s)$ [molar mass = $40\text{ g}$] D. $12\text{ g of } \text{H}_2(g)$ [molar mass = $2\text{ g}$] E. $220\text{ g of } \text{CO}_2(g)$ [molar mass = $44\text{ g}$] Choose the correct answer from the options given below:
A
A, B, and C only
B
A, B, and D only
C
B, C, and D only
D
B, D, and E only
The answer is B.
Solution
Total atoms $= n \times \text{atomicity} \times N_A$. A: $2 \times 6 = 12 N_A$; B: $4 \times 3 = 12 N_A$; D: $6 \times 2 = 12 N_A$. Thus A, B, and D contain equal numbers of atoms.
Calculating total moles of atoms: A. $\text{Na}_2\text{CO}_3$: Moles $= 212/106 = 2\text{ mol}$. Each unit contains 6 atoms, so total atoms $= 2 \times 6 N_A = 12 N_A$. B. $\text{Na}_2\text{O}$: Moles $= 248/62 = 4\text{ mol}$. Each unit contains 3 atoms, so total atoms $= 4 \times 3 N_A = 12 N_A$. C. $\text{NaOH}$: Moles $= 240/40 = 6\text{ mol}$. Each unit contains 3 atoms, so total atoms $= 6 \times 3 N_A = 18 N_A$. D. $\text{H}_2$: Moles $= 12/2 = 6\text{ mol}$. Each molecule has 2 atoms, so total atoms $= 6 \times 2 N_A = 12 N_A$. E. $\text{CO}_2$: Moles $= 220/44 = 5\text{ mol}$. Each molecule has 3 atoms, so total atoms $= 5 \times 3 N_A = 15 N_A$. Thus, A, B, and D each contain $12 N_A$ atoms.
ConceptCalculation of Number of Atoms
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