Question

tell me the answer of this question , please guys​

Answer 1

In sulphur estimation ,0.157 gm of an organic compound gave 0.4813 gm of BaSO4 , so we have to determine vthe percentage of sulphur in the compound.

• Atomic mass of Ba is 137 gm/mol

atomic mass of S is 32 gm/mol

atomic mass of O is 16 gm/mol

And the mass of the organic compound is 0.157 gm

Mass of Barium sulfate is 0.4813 gm

molecular mass of Barium sulfate= 137+32+(16×4) = 233 gm/mol

Now we can say that 233 gm of Barium sulfate contains 32 gm of sulphur

• 0.4813 gm BaSO4 or barium sulfate contains = 32/233 × 0.4813 gm sulphur=0.066 gm of sulphur

so the percentage of sulphur would be (0.066÷0.157)×100= 42.03%

Answer 2

Mass of Organic Comply =0.157g Mass of Baso4!0.4813g , Molecular Mass of Baso4=137+32+64=233 ,233g Baso4 contains 32g Sulphur, 0.4813 Baso4 would contain =32/233x0.4813 g sulphur =0.066g ,Percentage of 'S'=0.066/0.157x100=42.10%