Q70 (2026)
A student has been given $0.314\text{ g}$ of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained $0.4813\text{ g}$ of barium sulphate. The percentage of sulphur present in the compound is
(Given Molar mass in $\text{g mol}^{-1}$ $S:32$, $BaSO_{4}:233$)
- $(1)\text{ 42.10\%}$
- $(2)\text{ 63.15\%}$
- $(3)\text{ 21.05\%}$
- $(4)\text{ 48.24\%}$
### Related Formula
$$\%\text{ of S} = \frac{32}{233} \times \frac{\text{Mass of } BaSO_4}{\text{Mass of organic compound}} \times 100$$
### Core Logic
Mass of organic compound = $0.314\text{ g}$
Mass of $BaSO_4$ formed = $0.4813\text{ g}$
Substituting the values:
$$\%\text{ S} = \frac{32}{233} \times \frac{0.4813}{0.314} \times 100$$
$$\%\text{ S} = 0.1373 \times 1.5328 \times 100 \approx 21.052\%$$
### Step 1: Final Conclusion
The percentage of sulphur is approximately $21.05\%$.
### Pattern Recognition
Straight application of Carius method for Sulphur estimation formula.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques
