When 200 gm of an Oleum sample labeled as 109% is mixed with 300 gm of another Oleum sample labeled as 118%, the new labeling of resulting Oleum sample becomes -

Options

A. `112.6%`
B. `114.4%`
C. `113.5%`
D. `127%`

Answer

B

Explanation

After mixing 500 gm oleum sample consist of `320 gm SO_(3)` and `180 gm H_(2)SO_(4)`
100 gm oleum `rArr SO_(3) = (320)/(5) = 64 gm`
`SO_(3) + H_(2)O rarr H_(2)SO_(4)`
Moles `(64)/(80) " " (64)/(80)`
Mass of `H_(2)O = (64)/(80) xx 18 = 14.4 gm`
Labelling `= (100 + 14.4)%`
`114.4 %`