If 10gm of water4 is added to 150gm of oleum (104.5%), then the find solution:

Options

A. Mass of `SO_(3)` left is 10gm
B. Mass of `H_(2)SO_(4)` is 156.75gm
C. No water will be left
D. Labelling of new solution is 102.25%

Explanation

100gm convert into maximum 104.5gm `H_(2)SO_(4)`
150gm convert into pure `(104.5)/(100)xx150=156.75gm H_(2)SO_(4)`
and `w_(H_(2)O)` requried `=(4.5)/(100)xx150=6.75gm`
we add water `gt6.75gm`
thus no `SO_(3)` remain after 10gm water addition labelling of new solution `=100%` as no `SO_(3)` left in it.