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Q.

What is the % of free $S O_{3}$ in an oleum that is labelled as 104.5% $H_{2} S O_{4}$ ?

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a

10

b

20

c

40

d

30

answer is B.

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Detailed Solution

$S O_{3} + H_{2} O \to H_{2} S O_{4} n_{S O_{3}} = n_{H_{2} O} = n_{H_{2} S O_{4}} % o f f r e e S O_{3} = \frac{104.5 - 100}{18} \times 80 = 20 %$

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