Which of the gases contains the same number of molecules of $O_{2}$ as in 16 g of $O_{2}$ ?

16 g S O_{3}

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32 g S O_{2}

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16 g S O_{2}

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All of the above

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Solution

The correct option is B $32 g S O_{2}$
The molar mass of $O_{2} = 32$ g/mol.
Number of moles of $O_{2} = \frac{16 g}{32 g / / m o l} = 0.5 m o l$ .
Let N be Avogadro's number.
0.5 moles of $O_{2} = 0.5 N$ molecules of $O_{2}$ .
The molar mass of $S O_{2} = 64 g / m o l$
Number of moles of $S O_{2} = \frac{32 g}{64 g / / m o l} = 0.5 m o l$ .
0.5 moles of $S O_{2} = 0.5 N$ molecules of $S O_{2}$ .
Hence, 32 g $S O_{2}$ contains the same number of molecules of $O_{2}$ as in 16 g of $O_{2}$ .
Note:
Number of moles of 16 g $S O_{2} = \frac{16 g}{64 g / / m o l} = 0.25 m o l$ .
0.25 moles of $S O_{2} = 0.25 N$ molecules of $S O_{2}$ .
The molar mass of $S O_{3} = 80 g / m o l$
Number of moles of $S O_{3} = \frac{16 g}{80 g / / m o l} = 0.2 m o l$ .
0.2 moles of $S O_{3} = 0.2 N$ molecules of $S O_{3}$ .

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