31 gm `H_(2)SO_(4)` is mixed with 20 gram `SO_(3)` to form mixture
Determine % labelling of oleum solution

30 gm H_(2)SO_(4) is mixed with 20 gram SO_(3) to form mixture Determine % labelling of oleum solution

30 gm H_(2)SO_(4) is mixed with 20 gram SO_(3) to form mixture Find mole fraction of SO_(3)

30 gm H_(2)SO_(4) is mixed with 20 gram SO_(3) to form mixture Find mole fraction of SO_(3)

A mixutre is prepared by mixing 10 gm H_(2)SO_(4) and 40 gm SO_(3) calculate, (a) mole fraction of H_(2)SO_(4) (b) % labelling of oleum

An oleum sample contains 10 g SO_(3) and 15 g H_(2) SO_(4) Answer the following questions on the basis of above information : % labelling of oleum sample is .

An oleum sample contains 10 g SO_(3) and 15 g H_(2) SO_(4) Answer the following questions on the basis of above information : % labelling of oleum sample is .

An oleum sample contains 10 g SO_(3) and 15 g H_(2) SO_(4) Answer the following questions on the basis of above information : % labelling of oleum sample 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) ?

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 ' 019% 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)' ?

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 ' 019% 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) 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % free SO_(3) in the sample is : (a)74 (b)26 (c)20 (d)None of these