A `50 gm` oleum sample labelled as `104.5%` will have mass of `SO_(3)`

Options

A. `80 gm`
B. `40 gm`
C. `20 gm`
D. `10 gm`

Answer

D

Explanation

`104.5%` Labelled oleum sample
Moles of `H_(2)O= (4.5)/(18) = (1)/(4)` moles
`{:(,SO_(3),+,H_(2)O,rarr,H_(2)SO_(4)),("Moles",(1)/(4),,(1)/(4),,):}`
Mass of `SO_(3) = (1)/(4) xx 80 = 20 gm`
100 gm oleum has 20 gm `SO_(3)`
In 50 gm Oleum, `SO_(3)` will be 10 gm