Question

An oleum sample is labelled as (100 + x)% and it contain 90 % free SO3 by weight. Hence, x
is​

Answer 1

Answer:

118% oleum means 100 g of oleum requires 18 g of H

2

O to form 118 g of H

2

SO

4

.

SO

3

+H

2

O→H

2

SO

4

1 mol 1 mol

(80 g) (18 g)

18 g H

2

O ≡80 g SO

3

∴ Percentage of free SO

3

=

18

18

×80=80%