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Question

Percentage of free $$SO_3$$ in an oleum bottle labelled 113.5% $$H_2SO_4$$ is :


A
40
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B
60
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C
50
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D
45
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Solution

The correct option is B 60
We have
$$H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4$$

$$113.5 $$%$$ H_2SO_4$$ means 

100g of oleum conatins 113.5 g of $$H_2SO_4$$

so, amount of water reacted with oleum = 113.5 - 100 = 13.5g
moles of water = $$\dfrac{13.5}{18}= 0.75 moles$$

we also have,
$$H_2O + SO_3 \rightarrow H_2SO_4$$

so, 1 mole of water reacts with 1 mole of $$SO_3$$.

so, 0.75 moles of water reacts with 0.75 moles of $$SO_3$$

so, moles of $$SO_3$$ = 0.75

so, amount of $$SO_3$$ = $$moles \times molar\ mass = 0.75 \times 80 = 60g$$

Chemistry

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