Question

$0.4037$ g of an organic compound containing $S$ was treated with conc. $HN{O}_3$ in a Carius tube. On precipitation with $BaC{l}_2$ $0.1936$ g of $BaS{O}_4$ was $0.35$ g. Find the percentage of sulphur.

• A. $6.58$
• B. $7.12$
• C. $8.33$
• D. $1.67$

Answer 1

The correct option is A $6.58$

According to Carius method,

$\%$ of $S=\frac{32}{233}\times \frac{MassofBaS{O}_4}{Massoforganiccompound}\times 100$

$⟹\%$ of $S=\frac{32}{233}\times \frac{0.1936}{0.4037}\times 100$

$⟹\%$ of $S=13.7\times 0.479$

$⟹\%$ of $S=6.58\%$