Question

$0.53$ g of substance gave $1.158$ g of $BaS{O}_4$ by Carius method. Calculate the percentage of sulphur.

• A. $42.10$%
• B. $23.3$%
• C. $32.03$%
• D. $30$%

Answer 1

The correct option is D $30$%

Molar mass of $BaS{O}_4$ = 233 g

Suppose the mass of organic compound = m grams

Let the mass of barium sulfate formed = 1.158 grams

We know that 32 grams of sulfur is present in 1 mol of $BaS{O}_4$

Therefore, 233 g $BaS{O}_4$ contains 32 g sulfur

⇒ 1.158 g of $BaS{O}_4$ contains $\frac{32\times 1.158}{233}$ grams of sulphur

Percentage of sulphur = $\frac{32\times 1.158\times 100}{233\times 0.53}$ = 30%

RIGHT ANSWER IS (D).