In the estimation of sulphur by Carius method, 0.480 g of an organic compound gives 0.699 g of barium sulphate. The percentage of sulphur in this compound is (Atomic masses: Ba = 137, S = 32, 0 = 16)

Question

In the estimation of sulphur by Carius method, 0.480 g of an organic compound gives 0.699 g of barium sulphate. The percentage of sulphur in this compound is (Atomic masses: Ba = 137, S = 32, 0 = 16) (A) 20 % (B) 15 % (C) 35 % (D) 30 %

Tardigrade Admin · Accepted Answer

S % =(32/233) × ( text wt. of BaSO 4/ text wt. of organic compound ) × 100 =(32/233) × (0.699/0.480) × 100 =20 %

Q. In the estimation of sulphur by Carius method, $0.480 g$ of an organic compound gives $0.699 g$ of barium sulphate. The percentage of sulphur in this compound is (Atomic masses: $B a = 137, S = 32, 0 = 16)$

20 %

15 %

35 %

30 %

40 %

$S % = \frac{32}{233} \times \frac{wt. of BaSO _{4}}{wt. of organic compound} \times 100$

$= \frac{32}{233} \times \frac{0.699}{0.480} \times 100$

$= 20%$

Q. In the estimation of sulphur by Carius method, 0.480g of an organic compound gives 0.699g of barium sulphate. The percentage of sulphur in this compound is (Atomic masses: Ba=137,S=32,0=16)