Question

If the total number of electrons present in a pure sample of $FeS{O}_4.7H_{2}O$ is $4.336\times {10}^{26}$
then the total number of moles of oxygen atoms present in $FeS{O}_4.7H_{2}O$:

Answer 1

The total number of electrons in one formula unit of $FeS{O}_4.7H_{2}O$
$=(26+16+4\times 8+14\times 1+7\times 8)=144$

So, $144\times 6.022\times {10}^{23}$ electrons are present in 1 mole of $FeS{O}_4.7H_{2}O$.

Hence, $4.336\times {10}^{26}$ electrons will be present in
$=\frac{4.336\times {10}^{26}}{1.44\times 6.022\times {10}^{23}}\text{moles of}FeS{O}_4.7H_{2}O$

$=\text{5 moles of}FeS{O}_4.7H_{2}O$

From the formula unit :
1 mole of $FeS{O}_4.7H_{2}O$ contains 11 moles of $O$ atoms.

Thus, 5 moles of $FeS{O}_4.7H_{2}O$ will contains 55 moles of $O$ atoms.