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