$0.5952$ g of an organic substance, when treated by Carius method, gave $0.52$ g of $B a S O_{4}$ . Calculate the % of the compound.

23.16

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23.35

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11.96

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None of these

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Solution

The correct option is B $11.96$ %
Percentage of sulphur in the compound
= $\frac{32}{233} \times \frac{m a s s o f B a S O_{4}}{m a s s o f c o m p o u n d} \times 100$
$32 \times 0.52 \times \frac{100}{233 \times 0.5952} = 11.96$
C is answer

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