`2.0 g` of oleum is diluted with water. The solution was then neutralised by `432.5 mL of 0.1 N NaOH`. Select the correct statements:

Options

A. Equivalent of `H_(2)SO_(4)=0.03`
B. Equivalent of `SO_(3)=0.01325`
C. `%` of free `SO_(3)-26.5` in oleum
D. `%` of oleum `=108.11`

Answer

A, B, C, D

Explanation

`underset(a)(H_(2)SO_(4))+underset(b)(SO_(3))`
`:. a+b=12`....(i)
Also Meq.of `H_(2)SO_(4)+Meq.of SO_(3)="Meq.of" NaOH`
`(a)/(49)xx1000+(b)/(40)xx1000=0.1xx432.5`
`:. 40a+49b=84.77`
`:. a=1.47` and `b=0.53`
`:.` Equivalent of `H_(2)SO_(4)=(1.47)/(49)=0.03`
Now `SO_(3)+H_(2)OrarrH_(2)SO_(4)`
Equivalent of `SO_(3)=(0.53)/(40)=0.01325`
Wt.of `H_(2)O` to react with `SO_(3)`
`=(0.53xx18)/(80)=0.11925 g`
`:. 108.11% "oleum"=100 gH_(2)SO_(4)+8.11 g H_(2)O`
`=100gH_(2)SO_(4)+(8.11xx80)/(18)`
`100g H_(2)SO_(3)+36gSO_(3)`
`136 g"oleum"`
`136 g` oleum has `36 g SO_(3)`
`:. 2 g "oleum"=(36xx2)/(136)=0.53 g SO_(3)`
`:. %` of free `SO_(3)=(0.53xx100)/(2)=26.5`