Which of the following pairs have the same number of atoms?

16 g O_{2} (g)

and

4 g of H_{2} (g)

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16 g of O_{2} (g)

and

44 g of C O_{2} (g)

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28 g of N_{2} (g)

and

32 g of O_{2} (g)

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12 g of C (s)

and

23 g of N a (s)

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Solution

The correct option is D $12 g of C (s)$ and $23 g of N a (s)$
$16 g O_{2} = \frac{16}{32} = 0.5 mol = 2 \times 0.5 \times N_{A} atoms = N_{A} atoms$
$4 g H_{2} = \frac{4}{2} = 2 mol = 2 \times 2 \times N_{A} atoms = 4 N_{A} atoms$
$44 g C O_{2} = \frac{44}{44} = 1 mol = 3 \times 1 \times N_{A} atoms = 3 N_{A} atoms$
$28 g N_{2} = \frac{28}{28} = 1 mol = 2 \times 1 \times N_{A} atoms = 2 N_{A} atoms$
$32 g O_{2} = \frac{32}{32} = 1 mol = 1 \times 2 \times N_{A} atoms = 2 N_{A} atoms$
$12 g C = \frac{12}{12} = 1 mol = N_{A} atoms$
$23 g N a = \frac{23}{23} = 1 mol = N_{A} atoms$

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