Saved Bookmarks
| 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 -A. `112.6%`B. `114.4%`C. `113.5%`D. `127%` |
|
Answer» Correct Answer - B 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 %` |
|