In the estimation of sulphur by Carius method, 0.480 f og an organic comppund gives 0.699 g of barium sulphate. The percentage of sulphur in this compound is (atomic masses : Ba = 137, S = 32 and O = 16)

Options

A. 0.2
B. 0.15
C. 0.35
D. 0.3

Explanation

To find the percentage of sulfur in the organic compound using the Carius method, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- Weight of the organic compound (W_organic) = 0.480 g
- Weight of barium sulfate (W_BaSO4) = 0.699 g

2. **Calculate the molar mass of barium sulfate (BaSO4):**
- The formula for barium sulfate is BaSO4.
- Molar mass of Ba = 137 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (and there are 4 oxygen atoms)
- Therefore, the molar mass of BaSO4 = 137 + 32 + (16 × 4)
- Molar mass of BaSO4 = 137 + 32 + 64 = 233 g/mol

3. **Use the formula for the percentage of sulfur:**
- The percentage of sulfur in the organic compound can be calculated using the formula:
\[ \text{Percentage of S} = \left( \frac{\text{Molar mass of S}}{\text{Molar mass of BaSO4}} \right) \times \left( \frac{W_{BaSO4}}{W_{organic}} \right) \times 100 \]
- Substituting the values:
\[ \text{Percentage of S} = \left( \frac{32}{233} \right) \times \left( \frac{0.699}{0.480} \right) \times 100 \]

4. **Calculate the value:**
- First, calculate \( \frac{0.699}{0.480} \):
\[ \frac{0.699}{0.480} \approx 1.45625 \]
- Now, calculate \( \frac{32}{233} \):
\[ \frac{32}{233} \approx 0.137 \]
- Now, multiply these values:
\[ 0.137 \times 1.45625 \approx 0.200 \]
- Finally, multiply by 100 to get the percentage:
\[ 0.200 \times 100 = 20\% \]

5. **Conclusion:**
- The percentage of sulfur in the organic compound is **20%**.