Comprehension # 7 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 100g sample of oleum is diluted with desired weight of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109gtotal mass of pure H_(2)SO_(4) will be formed when 100g of oleum is diluted by 9g of H_(2)O which combines combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)OrarrH_(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 % of free SO_(3) in the sample is :

Comprehension # 7 Oleum is considered as a solution of SO_(3) in H_(2

1.	Comprehension # 7 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 100g sample of oleum is diluted with desired weight of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H_(2)SO_(4) means the 109gtotal mass of pure H_(2)SO_(4) will be formed when 100g of oleum is diluted by 9g of H_(2)O which combines combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)OrarrH_(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 % of free SO_(3) in the sample is :
Answer» `74` `26` `20` none of these Solution :`H_(2)SO_(4)+2NaOH rarr Na_(2)SO_(4)+2H_(2)O` `{:(,H_(2)SO_(4),+,2NaOHrarrNa_(2)SO_(4)+2H_(2)O),(,(0.4xx54)/(2xx1000),,0.4 N),(,,,54 ml):}` `wt`. Of `H_(2)SO_(4) = 1.058 gm` MOLE of `H_(2)O = (0.0584)/(18)` `n_(SO_(3)) = 0.003244` `wt`. of `SO_(3) = 0.26 gm` `%` free `SO_(3) = 26`