Question

What is the percentage of free so3 in oleum labeled as 109%h2so4?

Answer 1

This is the answer and you simplify it

Answer 2

Dear student,

Given :-

• oleum percentage = 109%

To find :-

• % free of SO₃

Solution :-

We know by the mole concept that, for % free SO₃, we have;

        % free SO₃ = 80/18 (y - 100)

where, y = percentage of oleum

Putting the given values in the above formula, we obtain;

 % free SO₃ = 80/ 18 (109 - 100)

                    = 80/ 18 × 9

                    = 80/2

                    = 40%

Hence, the % free SO₃ is 40.