0.102 g of an organic compound X was oxidized with fuming nitric acid. The resulting solution, after reaction with an excess of aqueous BaCl_2…

Question

0.102 g of an organic compound X was oxidized with fuming nitric acid. The resulting solution, after reaction with an excess of aqueous BaCl_2, produced 0.233 g of BaSO_4 as a precipitate. Compound X is likely to be : [Given: Atomic wt. of Ba=137]

MARKS Content Team · Accepted Answer

Correct answer: (D). 0.233 gm BaSO_4 has 1 millimole BaSO_4 and hence has lmillimole S ∴ organic compound X also has 1 millimole S % of S in 0.102 gm of organic compound X =0.032/0.102×100=31.37% 102 gm of this organic compound has 32 gm S This has same % of S

$0.102 g$ of an organic compound $X$ was oxidized with fuming nitric acid. The resulting solution, after reaction with an excess of aqueous ${BaCl}_{2}$ , produced $0.233 g$ of ${BaSO}_{4}$ as a precipitate. Compound $X$ is likely to be : [Given: Atomic wt. of $Ba = 137$ ]

Solution

More Some Basic Concepts of Chemistry Questions

0.102 g of an organic compound X was oxidized with fuming nitric acid. The resulting solution, after reaction with an excess of aqueous BaCl2, produced 0.233 g of BaSO4 as a precipitate. Compound X is likely to be : [Given: Atomic wt. of Ba=137]

Solution

More Some Basic Concepts of Chemistry Questions

$0.102 g$ of an organic compound $X$ was oxidized with fuming nitric acid. The resulting solution, after reaction with an excess of aqueous ${BaCl}_{2}$ , produced $0.233 g$ of ${BaSO}_{4}$ as a precipitate. Compound $X$ is likely to be : [Given: Atomic wt. of $Ba = 137$ ]