Comprehension # 6 The percentage labelling of oleum is a unique process by means of which, the percentagecomposition of H_(2)SO_(4), SO_(3) (free) and SO_(3) (combined) is calculated. Oleum is nothing but it is a mixture of H_(2)SO_(4) and SO_(3) i.e., H_(2)S_(2)O_(7),which is obtained by passing. SO_(3) in solution of H_(2)SO_(4). In order of dissolve free SO_(3) in oleum, dilution of oleum is done, in which oleum converts into pure H_(2)SO_(4). It is shown by the reaction as under : H_(2)SO_(4)+SO_(3)+H_(2)Orarr2H_(2)SO_(4)("pure") or ""SO_(3)+H_(2)OrarrH_(2)SO_(4)("pure") When 100g sample of oleum is diluted with desired weight of H_(2)O("in" g), then the total mass of pure H_(2)SO_(4) obtained after dilution is known as percentage labelling in oleum. For example, if the oleum sample is labelled as ""109%H_(2)SO_(4)" it means that 100 g of oleum on dilution with 9m of H_(2)O provides109g pure H_(2)SO_(4), in which all free SO_(2) in 100g of oleum is dissolved. What volume of 10 M NaOH ("in" mL) will be required to react completely with free SO_(3) in 118% labelled oleum sample?