$0.16 g$ of an organic compound containing sulphur produces $0.233 g$ of $B a S O_{4}$ . Percentage of sulphur in the compound is :

20

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80

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50

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10

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Solution

The correct option is A $20$
The molar masses of barium sulphate and sulphur are $233 g / m o l$ and $32.1 g / m o l$ respectively.

Number of moles of $B a S O_{4} = \frac{0.233}{233} = 0.001 m o l$ $= number of moles of S.$

Mass of $S = 0.001 \times 32.1 = 0.0321 g$

Percentage of $S = \frac{100 \times 0.0321}{0.16} = 20$ %

Hence, option A is correct.

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