The percentage labelling (mixture of `H_(2)SO_(4)` and `SO_(3))` refers to the total mass of pure `H_(2)SO_(4)`. The total amount of `H_(2)SO_(4)` found after adding calculated amount of water to `100 g` oleum is the percentage labelling of oleum. The higher the percentage lebeling of oleum higher is the amount of free `SO_(3)` in the oleum sample.
`100 g` sample of '149%' oleum was taken and calculated amount of `H_(2)O` was added to make `H_(2)SO_(4)`.`500 mL` solution of `x MKOH` solution is required to neutralize the solution. The value of `x` is.

Options

A. `1 M`
B. `2 M`
C. `4 M`
D. `6 M`

Answer

D

Explanation

147% oleum sample implies `100 g` oleum sample on addition of `47 g` of `H_(2) O` will form `147 g H_(2) SO_(4)`.
Equivalent of `H_(2) SO_(4) = (147)/(49) = 3 = 3 xx 10^(3) mEq`
`:. mEq "of" H_(2)SO_(4) =` mEq of `NaOH`
`3 xx 10^(3) = 500 mL xx x xx 1`
`x = (3000)/(500) = 6 M`