Question

In the estimation of sulphur by Carius method. x g of an organic compound gave 0.233 g of \(BaSO_{4}\) If the percentage of sulphur in it is 8.89%, the value of x is

• A. 0.12
• B. 0.24
• C. 0.36
• D. 0.48

Hint:

Always ensure you use the molar masses provided within the specific problem question, as they can sometimes vary slightly between exam papers.

Answer 1

The Correct Option is C

Step 1: Recall the Carius method.
Sulfur in an organic compound is turned into sulfate and weighed as barium sulfate $BaSO_4$. The percentage of sulfur uses how much sulfur sits inside that $BaSO_4$.
Step 2: Write the percentage formula.
\[ \%S=\frac{32}{233}\times\frac{\text{mass of }BaSO_4}{\text{mass of compound }x}\times100 \] Here 32 is the mass of sulfur and 233 is the molar mass of $BaSO_4$.
Step 3: Put in the known values.
With $\%S=8.89$ and mass of $BaSO_4=0.233$ g: \[ 8.89=\frac{32}{233}\times\frac{0.233}{x}\times100 \]
Step 4: Simplify the easy part.
Note $\frac{0.233}{233}=0.001$. So \[ 8.89=\frac{32\times0.001\times100}{x}=\frac{3.2}{x} \]
Step 5: Solve for x.
\[ x=\frac{3.2}{8.89}\approx0.36\ \text{g} \]
Step 6: State the answer.
\[ \boxed{x\approx0.36\ \text{g}} \]