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An oleum sample is labelled as $113.5 %$ . Identify the incorrect statement.

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The amount of free $S O_{3}$ in $50 g$ oleum sample is $30 g .$

The amount of $H_{2} S O_{4}$ in $50 g$ oleum sample is $30 g .$

The new labelling of oleum sample when $8 g .$ water is added in $100 g .$ original oleum sample is $(100 + \frac{137.5}{27}) %$

In the original $50 g$ oleum sample when $6.75 g$ water is added then $56.75 g H_{2} S O_{4}$ is produced.

answer is B.

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Detailed Solution

(a) $% free {SO}_{3} = \frac{13.5}{18} \times 80 = 60$

$∴ In 50 g oleum, 30 g {SO}_{3} is free.$

$5.09 g \leftarrow 100 g$

$% label = 105.09 = 100 + \frac{137.5}{27}$

(d) $50 + 6.75 = 56.75 g H_{2} {SO}_{4}$

Hence, option B is correct.

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