Question

Find the number of electrons present in a pure sample of $(N{H}_4{)}_{2}S{O}_4$ containing 8 moles of $O$-atoms?

• A. $6.43\times {10}^{25}$
• B. $8.43\times {10}^{25}$
• C. $5.34\times {10}^{25}$
• D. $8.34\times {10}^{25}$

Answer 1

The correct option is B $8.43\times {10}^{25}$

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 $O$-atomes 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 $(N{H}_4{)}_{2}S{O}_4$
$=70\times 6.022\times {10}^{23}\times 2=8.43\times {10}^{25}$