In an ore, the only oxidizable material is `Sn^(2+)`. This ore is titrated with a dichromate solution containing `2.5g` of `K_(2)Cr_(2)O_(7)` in `0.5 "litre"`. A `0.40 g` sample of the ore required `10.0 cm^(3)` of titrant to reach equivalence point. Calculate the percentage of tin in ore.

Answer

To calculate the percentage of tin in the ore, we can follow these steps:

### Step 1: Calculate the Normality of the Potassium Dichromate Solution
We start by calculating the normality of the potassium dichromate (K₂Cr₂O₇) solution.

1. **Given Data**:
- Mass of K₂Cr₂O₇ = 2.5 g
- Volume of solution = 0.5 L
- Molar mass of K₂Cr₂O₇ = 294 g/mol
- n-factor of K₂Cr₂O₇ = 6 (since it involves 2 Cr atoms, each changing from +6 to +3)

2. **Calculate Equivalent Weight**:
\[ \text{Equivalent weight of K₂Cr₂O₇} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{294 \text{ g/mol}}{6} = 49 \text{ g/equiv} \]

3. **Calculate Number of Equivalents**:
\[ \text{Number of equivalents} = \frac{\text{mass}}{\text{equivalent weight}} = \frac{2.5 \text{ g}}{49 \text{ g/equiv}} = 0.05102 \text{ equiv} \]

4. **Calculate Normality**:
\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume in L}} = \frac{0.05102 \text{ equiv}}{0.5 \text{ L}} = 0.10204 \text{ N} \]

### Step 2: Calculate the Number of Equivalents of K₂Cr₂O₇ Used in the Titration
The volume of the titrant used is 10 cm³ (or 0.01 L).

\[ \text{Number of equivalents of K₂Cr₂O₇} = \text{Normality} \times \text{Volume in L} = 0.10204 \text{ N} \times 0.01 \text{ L} = 0.0010204 \text{ equiv} \]

### Step 3: Relate the Equivalents of Tin (Sn²⁺) to K₂Cr₂O₇
From the stoichiometry of the reaction:
- 1 equivalent of K₂Cr₂O₇ reacts with 6 equivalents of Sn²⁺.
- Therefore, the equivalents of Sn²⁺ can be calculated as:
\[ \text{Equivalents of Sn²⁺} = 6 \times \text{Equivalents of K₂Cr₂O₇} = 6 \times 0.0010204 \text{ equiv} = 0.0061224 \text{ equiv} \]

### Step 4: Calculate the Weight of Sn²⁺
The equivalent weight of Sn²⁺ is:
\[ \text{Equivalent weight of Sn²⁺} = \frac{\text{Molar mass of Sn}}{n-factor} = \frac{118 \text{ g/mol}}{2} = 59 \text{ g/equiv} \]

Now, calculate the weight of Sn²⁺:
\[ \text{Weight of Sn²⁺} = \text{Equivalents} \times \text{Equivalent weight} = 0.0061224 \text{ equiv} \times 59 \text{ g/equiv} = 0.3602 \text{ g} \]

### Step 5: Calculate the Percentage of Tin in the Ore
The sample weight of the ore is 0.40 g.

\[ \text{Percentage of Sn in ore} = \left( \frac{\text{Weight of Sn²⁺}}{\text{Weight of ore}} \right) \times 100 = \left( \frac{0.3602 \text{ g}}{0.40 \text{ g}} \right) \times 100 = 90.05\% \]

### Final Answer
The percentage of tin in the ore is approximately **90.05%**.

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