Question

If we have a pure $CaC{O}_3$ sample that contains a total of $1.5055\times {10}^{22}$ electrons, then the number of millimoles of $CaC{O}_3$ present in the sample is:

• A. 0.2
• B. 0.3
• C. 0.03
• D. 0.5

Answer 1

The correct option is D 0.5

Total number of electrons present in 1 formula unit of $CaC{O}_3=20+6+(8\times 3)=50$

So, $50\times 6.022\times {10}^{23}$ electrons are present in 1 mol of $CaC{O}_3$.

Hence, the number of moles of $CaC{O}_3$ that would have $1.5055\times {10}^{22}$ electrons will be
$=\frac{1.5055\times {10}^{22}}{50\times 6.022\times {10}^{23}}=0.0005$
$=$ 0.5 millimoles