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)

100 ml of 0.3 M HCl is mixed with 200 ml of 0.3 M H_(2)SO_(4). Calculate the final molarity of the H^(+) ions in the resulting solution. Fill your answer by multiplying with 10.

Calculate the weight of 60% H_(2)SO_(4) required decomposing 50 g of chalk (Calcium carbonate ) .

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

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 :

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

Oleum or fuming sulphuric acid contains SO_(3) dissolved in sulphuric acid and has the molecular formula H_(2)S_(2)O_(7) , It is formed by passing SO_(3) in H_(2)SO_(4) . When water is added to oleum, SO_(3) reacts with water to form H_(2)SO_(4) . SO_(3)(g) + H_(2)O(l) to H_(2)SO_(4)(aq) As a result, mass of H_(2)SO_(4) increases. When 100 g sample of oleum is diluted with desired amount of water (in gram) then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling of oleum. % Labelling of oleum = Total mass of H_(2)SO_(4) present in oleum after dilution or = Mass of H_(2)SO_(4) initially present + Mass of H_(2)SO_(4) produced after dilution From this, the percentage composition of H_(2)SO_(4) and SO_(3) (free) and SO_(3) (combined) can be calculated. The percentage of SO_(3) in 109% H_(2)SO_(4) is

Oleum or fuming sulphuric acid contains SO_(3) dissolved in sulphuric acid and has the molecular formula H_(2)S_(2)O_(7) , It is formed by passing SO_(3) in H_(2)SO_(4) . When water is added to oleum, SO_(3) reacts with water to form H_(2)SO_(4) . SO_(3)(g) + H_(2)O(l) to H_(2)SO_(4)(aq) As a result, mass of H_(2)SO_(4) increases. When 100 g sample of oleum is diluted with desired amount of water (in gram) then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling of oleum. % Labelling of oleum = Total mass of H_(2)SO_(4) present in oleum after dilution or = Mass of H_(2)SO_(4) initially present + Mass of H_(2)SO_(4) produced after dilution From this, the percentage composition of H_(2)SO_(4) and SO_(3) (free) and SO_(3) (combined) can be calculated. The percentage of free SO_(3) and H_(2)SO_(4) in 112% H_(2)SO_(4) is