Question

A sample of $CaCO_3$ has 193000 C of charge in magniltide due to its anions only. Find the no. of moles of $CaCO_3$ in the sample. (mag. of charge on 1 mole $e^-$ = 96500 C)

• A. 1 mole
• B. 2 moles
• C. 0.5 moles
• D. 0.25 moles

Answer 1

Answer: (A)

1 mole

Answer 2

The anion in $CaCO_3$ is the carbonate ion, $CO_3^{2-}$. The magnitude of the charge on the carbonate ion is 2. The magnitude of charge on 1 mole of electrons is given as 96500 C (Faraday's constant, F). Therefore, the magnitude of charge on 1 mole of $CO_3^{2-}$ anions is $2 \times 96500$ C = 193000 C.

The total charge due to anions in the sample is given as 193000 C. Let $n$ be the number of moles of $CaCO_3$. Since one mole of $CaCO_3$ contains one mole of $CO_3^{2-}$ ions, 'n' moles of $CaCO_3$ contain 'n' moles of $CO_3^{2-}$ ions.

The total charge is calculated as: Total Charge = (Number of moles of anions) $\times$ (Charge per mole of anion) $193000 \text{ C} = n \times (193000 \text{ C/mol})$

Solving for $n$: $n = \frac{193000 \text{ C}}{193000 \text{ C/mol}}$ $n = 1 \text{ mole}$

Thus, the number of moles of $CaCO_3$ in the sample is 1 mole.