1.

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 -

Answer»

`112.6%`
`114.4%`
`113.5%`
`127%`

Solution :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 %`


Discussion

No Comment Found

Related InterviewSolutions