50 gm of `109 %` oelum is mixed with 50 gm of another `118 %` oleum. Calculate the maximum weight of `H_(2)SO_(4)` which can be obtained from the resulting mixture. (Fill the OMR after multiplying with 10)

What is the maximum mass of H_(2)SO_(4) that can be obtained from a 100 gm oleum sample labelled as 110%

Calculate the amount of 50% H_(2)SO_(4) required to decompose 25g calcium carbonate :-

25 gm of an oleum sample contains 15 gm H_(2)SO_(4) . What is % labellling of this oleum sample-

What weight of 98% pure H_(2)SO_(4) is required to neutralize 80gm of NaOH ?

When V ml of 2.2 M H_(2)SO_(4) solution is mixed with 10 V ml of water, the volume contraction of 2% take place. Calculate the molarity of diluted solution ?

If 50 gm oleum sample rated as 118% is mixed with 18 gm water, then the correct option is

A mixture is prepared by mixing 10 gm H_(2)SO_(4) and 40 gm SO_(3) calculate mole fraction of H_(2)SO_(4)

An oleum sample is labelled as 118%, Calculate Mass of H_(2)SO_(4) in 100 gm oleum sample

100 ml of 5 m H_2SO_4 of density 1 gm/ml is mixed with 100 ml of 8 m H_2SO_4 of density 1.25 g/mL. If there is no change in volume of resulting solution due to mixing, the molarity of the resulting mixture is -

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) What is the % of free SO_(3) in an oleum that is labelled as 104.5% H_(2)SO_(4) ?