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Oleum is considered as a solution of $$SO_3$$ in $$H_2SO_4$$, which is obtained by passing $$SO_3$$ in solution of $$H_2SO_4$$. When $$100\; g$$ sample of oleum is diluted with desired weight of $$H_2O$$, the total mass of $$H_2SO_4$$ will be for example, a oleum bottle labelled as $$109\%$$ $$H_2SO_4$$, means the $$109$$ g total mass of pure $$H_2SO_4$$ will be formed when $$100$$ g of oleum is diluted by $$9\; g$$ of $$H_2 O$$, which combines with all the free $$SO_3$$ to form $$H_2SO_4$$ as $$SO_3+H_2O \rightarrow H_2SO_4$$.

What is the $$\%$$ of free $$SO_3$$ in an oleum that is labelled as $$104.5\%$$ $$H_2SO_4$$?


A
10
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B
20
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C
40
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D
None of the above
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Solution

The correct option is A $$20$$
When $$100$$ g oleum sample requires $$4.5$$ g water to produce $$104.5$$ g sulfuric acid, it is labelled as $$104.5\%$$ oleum.
Free $$SO_3=\dfrac{80}{18}\times 4.5=20$$ g.
Percentage of free $$SO_3$$ present $$=\dfrac {20} {100} \times 100 =20\%$$.

Chemistry

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