Compound (X) has molar mass \(76\). \(2 \times 10^{-3}\) mole of X gives \(0.4813\,g\) \(BaSO_4\) as precipitate. What is the % of S in compound (X)? Mention answer in \(10^{-1}\) form.
- 215
- 325
- 435
- 515
The Correct Option is C
Solution and Explanation
Concept: Sulphur present in the compound is converted into \(BaSO_4\) precipitate. The relation between sulphur and \(BaSO_4\) is: \[ \frac{\text{Mass of S}}{\text{Mass of } BaSO_4} = \frac{32}{233} \] Thus sulphur percentage can be calculated from the mass of \(BaSO_4\).
Step 1: Mass of organic compound used \[ w = n \times M \] \[ w = 2 \times 10^{-3} \times 76 \] \[ w = 0.152\,g \]
Step 2: Mass relation between S and \(BaSO_4\) \[ \text{Mass of S} = \frac{32}{233} \times \text{mass of } BaSO_4 \]
Step 3: Calculate percentage of S \[ \%S = \frac{32}{233} \times \frac{0.4813}{0.152} \times 100 \] \[ \%S = 43.487 \]
Step 4: Convert to \(10^{-1}\) form
\[ 43.487 \times 10^{-1} = 434.87 \]
\[ \approx 435 \]
Thus,
\[ \boxed{435} \]
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