10 g of a mixture of `Cu_2S` and Cus was titrated with 200 " mL of " 0.75 M `MnO_4^(ɵ)` in acidic medium producing `SO_2,Cu^(2+)`, and `Mn^(2+)`. The `SO_2` was boiled off and the excess of `MnO_4^(ɵ)` was titrated with 175 " mL of " `1 M Fe^(2+)` solution. Find the percentage of CuS the in original mixture.

Explanation

To find the percentage of CuS in the original mixture of Cu2S and CuS, we can follow these steps:

### Step 1: Calculate the total moles of MnO4⁻ used in the reaction
The total volume of MnO4⁻ used is 200 mL, and its molarity is 0.75 M.

\[ \text{Total moles of MnO4⁻} = \text{Volume (L)} \times \text{Molarity} = 0.200 \, \text{L} \times 0.75 \, \text{mol/L} = 0.150 \, \text{mol} \]
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