`2.5 ` g sample of copper is dissolved in excess of `H_(2)SO_(4)` to prepare 100 mL of `0.02 M CuSO_(4)` (aq) . 10 mL of `0.02` M solution of `CuSO_(4)` (aq) is mixed with excess of Kl to show the following changes `CuSO_(4) + 2Kl to K_(2)SO_(4) + Cu l_(2)`
` 2Cu l_(2) to Cu_(2)l_(2) + l_(2)`
The liberated iodine is titrated with hypo `(na_(2)S_(2)O_(3))` and requires V mL of `0.1` M hypo solution for its complete reduction .
Percentage purity of sample is :

Options

A. `10.16`
B. `5.08`
C. `2.54`
D. `1.27`

Answer

To solve the problem step by step, we will follow the provided information and perform the necessary calculations.

### Step 1: Calculate the moles of CuSO₄ in the solution
We know that the concentration of the CuSO₄ solution is 0.02 M and the volume is 100 mL. First, we need to convert the volume from mL to L:

\[ \text{Volume in L} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]

Now, we can calculate the moles of CuSO₄:

\[ \text{Moles of CuSO₄} = \text{Molarity} \times \text{Volume in L} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} = 2 \, \text{mmol} \]

### Step 2: Relate moles of CuSO₄ to moles of copper
From the reaction, we see that 1 mole of CuSO₄ corresponds to 1 mole of copper. Therefore, the moles of copper in the sample is also:

\[ \text{Moles of Cu} = 2 \, \text{mmol} \]

### Step 3: Calculate the mass of copper
Now, we can calculate the mass of copper using its molar mass (63.5 g/mol):

\[ \text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar Mass of Cu} = 0.002 \, \text{mol} \times 63.5 \, \text{g/mol} = 0.127 \, \text{g} \]

### Step 4: Calculate the percentage purity of the copper sample
The percentage purity can be calculated using the formula:

\[ \text{Percentage Purity} = \left( \frac{\text{Mass of Cu}}{\text{Mass of Sample}} \right) \times 100 \]

Substituting the values we have:

\[ \text{Percentage Purity} = \left( \frac{0.127 \, \text{g}}{2.5 \, \text{g}} \right) \times 100 = 5.08\% \]

### Final Answer
The percentage purity of the copper sample is **5.08%**.

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