The percentage labelling (mixture of `H_(2)SO_(4)` and `SO_(3))` refers to the total mass of pure `H_(2)SO_(4)`. The total amount of `H_(2)SO_(4)` found after adding calculated amount of water to `100 g` oleum is the percentage labelling of oleum. The higher the percentage lebeling of oleum higher is the amount of free `SO_(3)` in the oleum sample.
What is the amount of free `SO_(3)` in an oleum sample labelled as '118%'.

Options

A. 0.4
B. 0.5
C. 0.7
D. 0.8

Answer

D

Explanation

118% oleum means `100 g` of oleum requires `18 g` of `H_(2) O` to form `118 g` of `H_(2)SO_(4)`
`underset({:(1 "mol"),((80 g)):})(SO_(3)) + underset({:(1 "mol"),((18 g)):})(H_(2)O) rarr H_(2) SO_(4)`
`:.` of free `SO_(3) = (18)/(18) xx 80 = 80%`