1.

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

Answer»

Mass of `SO_(3)` left is 10gm
Mass of `H_(2)SO_(4)` is 156.75gm
No water will be left
LABELLING of new solution is 102.25%

Solution :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.


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