An oleum sample is labelled as 113.5`%`. Identity the incorrect statement

200 gm of an oelum sample (labelled as 109 % ) is mixed with 400 gm of another oleum sample (labelled as 118 % ). The labelling of the new sample formed will be :

An oleum sample is labelled as (100+x)% and it contains 80/3% free SO_(3), by weight. Hence x is:

An oleum sample labelled as 104.5% in 10 g of this sample 90 mg water is added then which is/are correct for resulting solution.

An Oleum is labelled as 105%. Then 100 g of this Oleum will react with y gram of water. What will be value of y ?

An Oleum is labelled as 105%. Then 100 g of this Oleum will react with y gram of water. What will be value of y ?

100 gm oleum sample (labelled as X %) is mixed with excess water to make solution 4 litre. 1 L of this solution is neutralizes completely by (1)/(3) M,1.8 L NaOH solution. Calculate 10 X.