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 $H_{2}$

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Solution

c & d

The number of molecules can be calculated as follows

Number of molecules $= \frac{M a s s}{M o l a r m a s s} \times A v o g a d r o n u m b e r (N_{A})$

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 $O_{2} = 14 g o f N_{2} = 1.0 g o f H_{2}$

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