0.10 g of a sample containing `CuCO_(3)` and some inert impurity was dissolved in dilute sulphuric acid and volume made up to 520 mL. This solution was added into 50mL of 0.04 M KI solution where copper precipitates as Cul and `I^(-)` is oxidized into `I_(3)^(-)`. A 10 mL portion of this solution is taken for analysis, filtered and made up free `I_(3)^(-)` and then treated with excess of acidic permanganate solution. Liberated iodine required 20 mL of 2.5 mM sodium thiosulphate solution to reach the end point.
Determine mass percentage of `CuCO_(3)` in the original sample.

Options

A. 7.41
B. 74.1
C. 61.75
D. None of these

Answer

To determine the mass percentage of \( \text{CuCO}_3 \) in the original sample, we will follow these steps:

### Step 1: Calculate the total millimoles of \( \text{I}^- \) in the KI solution
Given:
- Volume of KI solution = 50 mL
- Molarity of KI solution = 0.04 M

Using the formula:
\[ \text{Millimoles of } I^- = \text{Volume (mL)} \times \text{Molarity (mol/L)} \]
\[ \text{Millimoles of } I^- = 50 \, \text{mL} \times 0.04 \, \text{mol/L} = 2 \, \text{mmol} \]

### Step 2: Calculate the milliequivalents of sodium thiosulphate used
Given:
- Volume of sodium thiosulphate solution = 20 mL
- Concentration of sodium thiosulphate = 2.5 mM

Using the formula:
\[ \text{Milliequivalents of } Na_2S_2O_3 = \text{Volume (mL)} \times \text{Concentration (mol/L)} \]
\[ \text{Milliequivalents of } Na_2S_2O_3 = 20 \, \text{mL} \times 2.5 \times 10^{-3} \, \text{mol/L} = 0.05 \, \text{mmol} \]

### Step 3: Calculate the millimoles of \( I^- \) that reacted
The total millimoles of \( I^- \) initially present = 2 mmol.
The millimoles of \( I^- \) that reacted with sodium thiosulphate = 0.05 mmol.

Thus, the millimoles of \( I^- \) that reacted with copper:
\[ \text{Millimoles of } I^- \text{ that reacted with Cu} = 2 \, \text{mmol} - 0.05 \, \text{mmol} = 1.95 \, \text{mmol} \]

### Step 4: Use stoichiometry to find millimoles of \( \text{Cu}^{2+} \)
From the reaction:
\[ 2 \, \text{Cu}^{2+} + 5 \, I^- \rightarrow 2 \, \text{CuI} + I_3^- \]
The stoichiometry shows that 2 moles of \( \text{Cu}^{2+} \) react with 5 moles of \( I^- \).

Using the ratio:
\[ \frac{\text{moles of } Cu^{2+}}{\text{moles of } I^-} = \frac{2}{5} \]
Thus,
\[ \text{Millimoles of } Cu^{2+} = \frac{2}{5} \times 1.95 \, \text{mmol} = 0.78 \, \text{mmol} \]

### Step 5: Calculate the mass of \( \text{CuCO}_3 \)
The molar mass of \( \text{CuCO}_3 \) is approximately 123.5 g/mol.

Using the formula:
\[ \text{Mass of } CuCO_3 = \text{Millimoles} \times \text{Molar Mass} \]
\[ \text{Mass of } CuCO_3 = 0.78 \, \text{mmol} \times 123.5 \, \text{g/mol} = 0.0963 \, \text{g} \]

### Step 6: Calculate the mass percentage of \( \text{CuCO}_3 \) in the original sample
Given the total mass of the sample is 0.10 g:
\[ \text{Mass percentage of } CuCO_3 = \left( \frac{\text{Mass of } CuCO_3}{\text{Total mass of sample}} \right) \times 100 \]
\[ \text{Mass percentage of } CuCO_3 = \left( \frac{0.0963 \, \text{g}}{0.10 \, \text{g}} \right) \times 100 = 96.3\% \]

### Final Answer
The mass percentage of \( \text{CuCO}_3 \) in the original sample is **96.3%**.

---