A pure sample of $(N H_{4})_{2} S O_{4}$ contains $8$ moles of $O$ -atoms. Find the number of electrons present in the sample.

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Solution

From the formula unit of $(N H_{4})_{2} S O_{4}$ , $4$ moles of $O$ -atoms are present in 1 mole of $(N H_{4})_{2} S O_{4}$ .

Therefore, $8$ moles of $O -$ atoms will be present in
$= \frac{8}{4}$ moles of $(N H_{4})_{2} S O_{4}$

Again, the total number of electrons in one formula unit of $(N H_{4})_{2} S O_{4}$
$= 7 \times 2 + 8 \times 1 + 16 + 8 \times 4 = 70$

So, the total number of electrons in one mole of $(N H_{4})_{2} S O_{4}$
$= 70 \times 6.022 \times 10^{23}$

Therefore, the total number of electrons in $\frac{8}{4}$ moles of $(N H_{4})_{2} S O_{4}$
$= 70 \times 6.022 \times 10^{23} \times 2 = 8.43 \times 10^{25}$

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