For 109% labeled oleum if the number of moles of H2SO4 and free SO3 be x and y respectively,then what will be the value of x+yx−y?
A
1
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B
18
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C
13
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D
9.9
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Solution
The correct option is D9.9 Reaction involved SO3+H2O→H2SO4
According to the stochiomentry of the reaction
18 g H2O will react with 80 g of SO3 to give 98 g of H2SO4.
So by unitary method in 109% oleum, 9 g of H2O will react with 40 g of SO3 to give 49 g of H2SO4.
This implies that in 100 g oleum, 40 g (free) SO3 and 60 g H2SO4 are present.
Hence moles of H2SO4 present (x) = 6098=0.6122mol
Similarly moles of SO3(y) = 4080=0.5mol
Hence the value of x−yx+y =0.6122+0.50.6122−0.5=9.91