60 gr of oleum (labelled as 118%) is mixed with 11.8 gm of water. What will be that composition of final mixture?

Options

A. Only `H_(2)SO_(4)`, having mass 71.8 gm
B. 118 gm of `H_(2)SO_(4)`
C. 70.8 gm `H_(2)SO_(4)` & 1 gm water
D. 32 gm `SO_(3)` & 39.8 gm `H_(2)SO_(4)`

Answer

3

Explanation

100 gm will contain 80 gm `SO_(3)`
60 gm will contain `80/100xx60=48 gm `
`{:(SO_(3),+,H_(2)O,to,H_(2)SO_(4)),(48 gm,,10.8 gm ,,58.8 gm),(0.6"mole",,"0.6 mole",,),(,,10.8" mole",,):}`
water left =11.8-10.8=1 gm
`H_(2)SO_(45)=58.8+12=70.8 gm `