109% labelled oleum has x mole of `H_(2)SO_(4)` and y mole of `SO_3` respectively. What is the value of (x+y)/(x-y)

Options

A. `8.81`
B. `9.91`
C. `10.6`
D. `11.6`

Answer

B

Explanation

`109%` oleum means it has `9 g H _(2)O` which reacts with free `SO _(3)` to given `H _(2) SO _(4)`
According to stoichiometry `SO _(3) + H _(2) O to H _(2) SO _(4)`
Mass of `SO _(3)` which will react with `9 g H _(2) O = ( 80 xx 9)/(18) = 40 g`
Let 100 g sample of `109%` labelled oleum contain 40 g `SO _(3) and 60 g H _(2) SO _(4)`
`therefore` Mole of `H _(2) SO _(4) = ( 60)/(98) = 0. 6122 = x ` moles of `SO _(3) = ( 40)/(80) = 0.5 = y`
`x + y = 1.112 x - y = 0. 1122`
`(x + y )/( x -y) = (1.1122)/(0.1122)= 9.91`