Question

If we have a pure $N{a}_{2}S{O}_4$ sample that contains a total of 7 billion electrons, then find the number of moles of $N{a}_{2}S{O}_4$ present in the sample.

• A. $1.66\times {10}^{-16}$ moles
• B. $1.66\times {10}^{-14}$ moles
• C. $3.66\times {10}^{-16}$ moles
• D. $3.66\times {10}^{-14}$ moles

Answer 1

The correct option is A $1.66\times {10}^{-16}$ moles

One molecule of $N{a}_{2}S{O}_4$ contains = $(11\times 2)+16+(8\times 4)=70$ electrons.
So, the number of electrons in 1 mole of $N{a}_{2}S{O}_4=70\times 6.022\times {10}^{23}$.
Therefore, $7\times {10}^9$ electrons will be present in
$=\frac{7\times {10}^9}{70\times 6.022\times {10}^{23}}$ moles of $N{a}_{2}S{O}_4$
$=1.66\times {10}^{-16}$ moles of $N{a}_{2}S{O}_4$