1.

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

Answer»

`80 gm`
`40 gm`
`20 gm`
`10 gm`

Solution :`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


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