Question

If the total mass of oxygen present in $FeS{O}_4.7H_{2}O$ is $176\text{g}$ the total number of electrons in $FeS{O}_4.7H_{2}O$ is $y\times {10}^{25}$, then the value of $y$ is:
(Given, $N_A=6\times {10}^{23}$)

Answer 1

We know, $176\text{g}$ of O atom = $\frac{176}{16}$ mol of $O$ atoms = 11 mol of $O$ atoms
Again, from the formula unit of, $FeS{O}_4.7H_{2}O$
11 $O$ atoms are present in one molecule of $FeS{O}_4.7H_{2}O$
So, 11 mol of $O$ atoms are present in 1 mol of $FeS{O}_4.7H_{2}O$
Number of electrons in one formula unit of $FeS{O}_4.7H_{2}O=26+16+11\times 8+14\times 1=144$
Therefore, number of electrons in 1 mol of $FeS{O}_4.7H_{2}O$ $=144\times 6\times {10}^{23}=8.64\times {10}^{25}$