Question

The total number of moles of electrons present in $12.044\times {10}^{14}P{O}_4^{3-}$ ions is ${10}^{-x}$. x is

Answer 1

We know,
$12.044\times {10}^{14}P{O}_4^{3-}$ ions = $\frac{12.0444\times {10}^{14}}{6.022\times {10}^{23}}$ moles $=2\times {10}^{-9}$ moles of $P{O}_4^{3-}$ ions.
The number of electrons in one $P{O}_4^{3-}$ ion $=15+(8\times 4)+3=50$ electrons.
So, the number of moles of electrons in one mole of $P{O}_4^{3-}$ ion $=50$ moles
Therefore, the total number of moles of electrons in $2\times {10}^{-9}$moles of $P{O}_4^{3-}$ ions
$=50\times 2\times {10}^{-9}$ moles $={10}^{-7}$ moles