Question

$0.16$ g of an organic substance was heated in a Carius tube and sulphuric acid was precipitated as $BaS{O}_4$ with $BaC{l}_2$.The weight of dry $BaS{O}_4$ was $0.35$ g. Find the percentage of sulphur.

• A. $32$%
• B. $25$%
• C. $28$%
• D. $30.04$%

Answer 1

Answer: (D)

$30.04$%

Total mass of organic compound = $0.16g$ [Given]

Mass of barium sulphate formed = $0.350g$ [Given]

1 mol of $BaS{O}_4$ = $233g$ = $32g$ of sulphur

Thus, $0.350g$ of $BaS{O}_4$ contains sulphur = $\frac{0.35\times 32}{233}g$=$=0.048g$

Therefore, percentage of sulphur =$\frac{0.048}{0.16}\times 100=30.04$

Option D is correct.