Question

Find the number of electrons present in a pure sample of $MgS{O}_4.7H_{2}O$ containing 5.5 moles of oxygen atoms?
(Take Avogadro's number $\left(N_A\right)=6\times {10}^{23}$)

• A. $3.9\times {10}^{25}$
• B. $4.2\times {10}^{23}$
• C. $5.8\times {10}^{25}$
• D. $4.5\times {10}^{23}$

Answer 1

The correct option is A $3.9\times {10}^{25}$

From the formula unit of ($MgS{O}_4.7H_{2}O$), 11 moles of $O$ atoms are present in 1 mole of $MgS{O}_4.7H_{2}O$.

Therefore, 5.5 moles of $O$ atoms will be present in
$=\frac{5.5}{11}$ moles of $MgS{O}_4.7H_{2}O$

Again, the total number of electrons in one formula unit of $MgS{O}_4.7H_{2}O$
$=1\times 12+1\times 16+14+11\times 8=130$

So, the total number of electrons in one mole of $MgS{O}_4.7H_{2}O$
$=130\times 6\times {10}^{23}$

Therefore, the total number of electrons in $\frac{5.5}{11}$ moles of $MgS{O}_4.7H_{2}O$
$=130\times 6\times {10}^{23}\times \frac{1}{2}=3.9\times {10}^{25}$