Question

If we have a pure $N{a}_{2}C{O}_3$ sample that contains a total of $7.8286\times {10}^{24}$ electrons, then find the number of moles of $N{a}_{2}C{O}_3$ present in the sample.

• A. 0.5 moles
• B. 0.3 moles
• C. 0.7 moles
• D. 0.25 moles

Answer 1

The correct option is D 0.25 moles

Total number of electrons present in one molecule of $N{a}_{2}C{O}_3=(11\times 2)+6+(3\times 8)=52$.
So, $52\times 6.022\times {10}^{24}$ electrons are present in 1 mole of $N{a}_{2}C{O}_3$
Therefore, $7.8286\times {10}^{24}$ electrons will be present in
$=\frac{7.8286\times {10}^{24}}{52\times 6.022\times {10}^{23}}$ moles of $N{a}_{2}C{O}_3$
$=$ 0.25 moles of $N{a}_{2}C{O}_3$