Question

16 g of oxygen has same number of molecules as in

(a) 16 g of CO (b) 28 g of $N_2$ (c) 14 g of $N_2$ (d) 1.0 g of $H_2$

Answer 1

c & d

The number of molecules can be calculated as follows

Number of molecules $=\frac{Mass}{Molarmass}\times Avogadronumber\left(N_A\right)$

Number of molecules, in 16 g oxygen $=\frac{16}{32}\times N_A=\frac{N_A}{2}$

In 16 g of CO $=\frac{16}{28}\times N_A=\frac{N_A}{1.75}$

In 28 g of $N_2=\frac{28}{28}\times N_A=N_A$

In 14 g of $N_2=\frac{14}{28}\times N_A=\frac{N_A}{2}$

In 1 g of $H_2=\frac{1}{2}\times N_A=\frac{N_A}{2}$

So, 16 g of $O_2=14gof{N}_2=1.0gof{H}_2$