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

(a) 16 g of $O_{2} (g)$ and 4 g of $H_{2} (g)$

(b) 16 g of $O_{2}$ and 44 g of $C O_{2}$

(c) 28 g of $N_{2}$ and 32 g of $O_{2}$

(d) 12 g of C(x) and 23 g of Na(s)

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Solution

(c) & (d)

Number of atoms in 28 g of $N_{2} = \frac{28}{28} \times N_{A} \times 2 = 2 N_{A}$ (where, $N_{A}$ = Avogadro number)

Number of atoms in 32 g of $O_{2} = \frac{32}{32} \times N_{A} \times 2 = 2 N_{A}$

12 g of C(s) contains atoms $= \frac{12}{12} \times N_{A} \times 1 = N_{A}$

Number of atoms in 23 g of $N a (s) = \frac{23}{23} \times N_{A} \times 1 = N_{A}$

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