Question

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

(a) 16 g of $O_2\left(g\right)$ and 4 g of $H_2\left(g\right)$

(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)

Answer 1

(c) & (d)

Number of atoms in 28 g of $N_2=\frac{28}{28}\times N_A\times 2=2N_A$ (where, $N_A$ = Avogadro number)

Number of atoms in 32 g of $O_2=\frac{32}{32}\times N_A\times 2=2N_A$

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

Number of atoms in 23 g of $Na\left(s\right)=\frac{23}{23}\times N_A\times 1=N_A$