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 $B a = 137, S = 32, O = 16$ )

20

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Solution

The correct option is E $20$ %
$S$ % $= \frac{32}{233} \times \frac{w t . o f B a S O_{4}}{w t . o f o r g a n i c c o m p o u n d} \times 100$
$= \frac{32}{233} \times \frac{0.699}{0.480} \times 100$
$= 20$ %

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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,O=16)

The correct option is E 20%S% =32233×wt.ofBaSO4wt.oforganiccompound×100 =32233×0.6990.480×100 =20%

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 $B a = 137, S = 32, O = 16$ )

The correct option is E $20$ %
$S$ % $= \frac{32}{233} \times \frac{w t . o f B a S O_{4}}{w t . o f o r g a n i c c o m p o u n d} \times 100$
$= \frac{32}{233} \times \frac{0.699}{0.480} \times 100$
$= 20$ %