How much of sulphur is present in an organic compound if 0.53g of compound gave 1.158g of baso4 on analysis
Answers
Answered by
1
Good evening,
we know, mass of organic compound=0.53 grams
Mass of barium sulphate formed=1.158 grams
32 grams of sulphur is present in 1 mol of BaSO4.
Molar mass of BaSO4=233 grams
therefore, 233 grams of BaSO4 contains 32 grams of sulphur
Therefore, 1.158 grams of BaSO4 contains
=(32*1.158)/233
Therefore, % of sulphur in it
=(32*1.158)/233*0.53 ×100
=30%
Have any queries, comment them down
^_^
we know, mass of organic compound=0.53 grams
Mass of barium sulphate formed=1.158 grams
32 grams of sulphur is present in 1 mol of BaSO4.
Molar mass of BaSO4=233 grams
therefore, 233 grams of BaSO4 contains 32 grams of sulphur
Therefore, 1.158 grams of BaSO4 contains
=(32*1.158)/233
Therefore, % of sulphur in it
=(32*1.158)/233*0.53 ×100
=30%
Have any queries, comment them down
^_^
Answered by
1
we know, mass of organic compound=0.53 grams
Mass of barium sulphate formed=1.158 grams
32 grams of sulphur is present in 1 mol of BaSO4.
Molar mass of BaSO4=233 grams
therefore, 233 grams of BaSO4 contains 32 grams of sulphur
Therefore, 1.158 grams of BaSO4 contains
=(32*1.158)/233
Therefore, % of sulphur in it
=(32*1.158)/233*0.53 ×100
Mass of barium sulphate formed=1.158 grams
32 grams of sulphur is present in 1 mol of BaSO4.
Molar mass of BaSO4=233 grams
therefore, 233 grams of BaSO4 contains 32 grams of sulphur
Therefore, 1.158 grams of BaSO4 contains
=(32*1.158)/233
Therefore, % of sulphur in it
=(32*1.158)/233*0.53 ×100
Similar questions