Question

Total charge carried by all the $\text{Ca}^{2+}$ ions present in 518 mL of 0.01 N $\text{CaCl}_2$ solution is n × $10^2$ coulombs. What is value of n? (Nearest integer)

Answer 1

5

Answer 2

To find the total charge carried by all the $\text{Ca}^{2+}$ ions, we need to determine the number of moles of $\text{Ca}^{2+}$ ions and then multiply by the charge per mole of $\text{Ca}^{2+}$ ions.

1. Calculate the molarity of $\text{CaCl}_2$ solution: Normality (N) is related to Molarity (M) by the valency factor (z). For $\text{CaCl}_2$, it dissociates as $\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^{-}$. The valency factor (z) for $\text{CaCl}_2$ (in terms of $\text{Ca}^{2+}$ ions contributing to charge) is 2. $\text{Molarity (M)} = \frac{\text{Normality (N)}}{\text{z}}$ $\text{M} = \frac{0.01 \text{ N}}{2} = 0.005 \text{ M}$

2. Calculate the moles of $\text{CaCl}_2$ in the given volume: Volume of solution = 518 mL = 0.518 L $\text{Moles of CaCl}_2 = \text{Molarity} \times \text{Volume (in L)}$ $\text{Moles of CaCl}_2 = 0.005 \text{ mol/L} \times 0.518 \text{ L} = 0.00259 \text{ mol}$

3. Calculate the moles of $\text{Ca}^{2+}$ ions: From the dissociation equation $\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^{-}$, 1 mole of $\text{CaCl}_2$ produces 1 mole of $\text{Ca}^{2+}$ ions. So, $\text{Moles of Ca}^{2+} = 0.00259 \text{ mol}$

4. Calculate the total charge carried by $\text{Ca}^{2+}$ ions: Each $\text{Ca}^{2+}$ ion carries a charge of +2 elementary charges. The charge carried by one mole of $\text{Ca}^{2+}$ ions is 2 Faradays (2F). 1 Faraday (F) = 96485 C/mol (Coulombs per mole of elementary charge). $\text{Total charge} = \text{Moles of Ca}^{2+} \times (\text{Charge per mole of Ca}^{2+})$ $\text{Total charge} = 0.00259 \text{ mol} \times (2 \times 96485 \text{ C/mol})$ $\text{Total charge} = 0.00259 \times 192970 \text{ C}$ $\text{Total charge} = 500.8923 \text{ C}$

5. Express the charge in the required format and find n: The total charge is given as $\text{n} \times 10^2$ coulombs. $\text{n} \times 10^2 = 500.8923$ $\text{n} = \frac{500.8923}{100}$ $\text{n} = 5.008923$

6. Round n to the nearest integer: The nearest integer to 5.008923 is 5.

The final answer is 5.

Explanation of the solution:

1. Convert Normality to Molarity: For $\text{CaCl}_2$, the valency factor is 2. Molarity = Normality/2 = 0.01/2 = 0.005 M.
2. Calculate Moles of $\text{CaCl}_2$: Moles = Molarity × Volume (in L) = 0.005 M × 0.518 L = 0.00259 mol.
3. Determine Moles of $\text{Ca}^{2+}$: Since 1 mole of $\text{CaCl}_2$ yields 1 mole of $\text{Ca}^{2+}$, moles of $\text{Ca}^{2+}$ = 0.00259 mol.
4. Calculate Total Charge: Each mole of $\text{Ca}^{2+}$ carries a charge of 2 Faradays (2F). Using F = 96485 C/mol, Total charge = 0.00259 mol × (2 × 96485 C/mol) = 500.8923 C.
5. Find n: Expressing the charge as n × 10² C, we get n × 100 = 500.8923, so n = 5.008923.
6. Nearest Integer: The nearest integer value for n is 5.