Question

When 200g of an oleum sample labelled as 109% is mixed with 300gm of another oleum sample labelled as 118% the new labelling of resulting oleum sample becomes

Answer 1

Answer:

200 gm of an oelum sample (labelled as 109) is mixed with 400 gm of another oleum sample (labelled as 118). The labelling of the new sample formed will be :

Answer 2

Oleum is mixture of H2SO4 and free SO3.

Labeling of oleum is based on the total amount of H2SO4 formed after reacting 100 g oleum with just enough H2O. The additional mass comes form amount of H2O required to convert all SO3 present in 100 g oleum into complete H2SO4.

Reaction involved is SO3 + H2O = H2SO4.

Here, 18 g H2O is required for 80 g SO3.

So, in 109 % oleum, 9 g H2O would be required for 40 g SO3.

It simply means that in 100 g oleum, 40 g (free) SO3 and 60 g H2SO4 are present.

So, in 100 g Oleum, number of moles of H2SO4 = (60/98) = 0.612 = X &

number of moles of SO3 = (40/80) = 0.5 = Y.

(X + Y) = 1.112

(X - Y) = 0.112

So, (X + Y)/ ( X-Y) = 1.112/ 0.112 = 9.92 = 10 (approx.)