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Arrhenius Acid

0.16 g of an ...

Question

$0.16$ g of an organic substance was heated in a Carius tube and sulphuric acid was precipitated as $B a S O_{4}$ with $B a C l_{2}$ .The weight of dry $B a S O_{4}$ was $0.35$ g. Find the percentage of sulphur.

32

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25

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28

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30.04

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Solution

The correct option is C $30.04$ %
Total mass of organic compound = $0.16 g$ [Given]

Mass of barium sulphate formed = $0.350 g$ [Given]

1 mol of $B a S O_{4}$ = $233 g$ = $32 g$ of sulphur

Thus, $0.350 g$ of $B a S O_{4}$ contains sulphur = $\frac{0.35 \times 32}{233} g$ = $= 0.048 g$

Therefore, percentage of sulphur = $\frac{0.048}{0.16} \times 100 = 30.04$
Option D is correct.

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