638.0 g of `CuSO_(4)` solution is titrated with excess of 0.2 M KI solution. The liberated `I_(2)` required 400 " mL of " 1.0 M `Na_(2)S_(2)O_(3)` for complete reaction. The percentage purity of `CuSO_(4)` in the sample is

Options

A. `5%`
B. `10%`
C. `15%`
D. `20%`

Explanation

To find the percentage purity of `CuSO4` in the solution, we will follow these steps:

### Step 1: Determine the moles of `Na2S2O3` used
Given that 400 mL of 1.0 M `Na2S2O3` is used, we can calculate the moles of `Na2S2O3`:

\[ \text{Moles of } Na2S2O3 = \text{Volume (L)} \times \text{Molarity} = 0.400 \, \text{L} \times 1.0 \, \text{mol/L} = 0.400 \, \text{mol} \]

### Step 2: Relate moles of `Na2S2O3` to moles of `I2`
The reaction between `I2` and `Na2S2O3` is as follows:

\[ I2 + 2 \, Na2S2O3 \rightarrow 2 \, NaI + Na2S4O6 \]

From this reaction, we see that 1 mole of `I2` reacts with 2 moles of `Na2S2O3`. Therefore, the moles of `I2` produced can be calculated as:

\[ \text{Moles of } I2 = \frac{0.400 \, \text{mol}}{2} = 0.200 \, \text{mol} \]

### Step 3: Relate moles of `I2` to moles of `CuSO4`
From the overall reaction of `CuSO4` with `KI`, we can establish the stoichiometry:

\[ 2 \, CuSO4 + 4 \, KI \rightarrow Cu2I2 + 2 \, K2SO4 + I2 \]

From this reaction, we see that 2 moles of `CuSO4` produce 1 mole of `I2`. Therefore, the moles of `CuSO4` can be calculated as:

\[ \text{Moles of } CuSO4 = 2 \times \text{Moles of } I2 = 2 \times 0.200 \, \text{mol} = 0.400 \, \text{mol} \]

### Step 4: Calculate the mass of `CuSO4`
The molar mass of `CuSO4` is approximately 159.5 g/mol. Hence, the mass of `CuSO4` in the solution can be calculated as:

\[ \text{Mass of } CuSO4 = \text{Moles} \times \text{Molar Mass} = 0.400 \, \text{mol} \times 159.5 \, \text{g/mol} = 63.8 \, \text{g} \]

### Step 5: Calculate the percentage purity of `CuSO4`
Now, we can find the percentage purity of `CuSO4` in the original solution:

\[ \text{Percentage Purity} = \left( \frac{\text{Mass of } CuSO4}{\text{Total Mass of Solution}} \right) \times 100 = \left( \frac{63.8 \, \text{g}}{638.0 \, \text{g}} \right) \times 100 \approx 10.0\% \]

### Final Answer
The percentage purity of `CuSO4` in the sample is approximately **10.0%**.

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