Question

$0.61$ g of an organic substance was heated in carius tube and the sulphuric acid formed was precipitated as $BaS{O}_4$ with $BaC{l}_2$. The weight of the dry $BaS{O}_4$ was $0.53$ g. Find the percentage of Sulphur.

• A. $11.90$%
• B. $43$%
• C. $23.04$%
• D. $51.2$%

Answer 1

Answer: (A)

$11.90$%

Mass of organic compound = 0.61 g

Mass of = 0.53 g

Molar mass of = 233 g/mole

Molar mass of sulfur = 32 g/mol

First we have to calculate the mass of sulfur.

As 233 g of contains = 32 g of sulfur

So, 0.53 g of contains = $\frac{32}{233}\times 0.53$ = 0.072 of sulfur

Now we have to calculate the percentage of sulfur in an organic compound.

= $\frac{0.072}{61}\times 100$

= 11.90