Question

A sample of oleum is labelled 109%. The % of free $S{O}_3$ in the sample is :

• A. 40%
• B. 80%
• C. 60%
• D. 9%

Answer 1

The correct option is A 40%

Oleum is a combination of sulphuric acid $+$ sulphuric trioxide and free $S{O}_3$ in oleum on reaction with $H_{2}O⟶H_{2}S{O}_4$ .

$S{O}_3+H_{2}O⟶H_{2}S{O}_4⟶\left(1\right)$

Let us take $109$% sample has $100gm$ of oleum and we added $9gms$ of water.

By reaction $\left(1\right)$ $18gms$ of water $⟶80gm$

$9gm$ of water $⟶40gm$

As per our assumption $40gm=40$% .