InterviewSolution
Saved Bookmarks
| 1. |
When `250 mg` of eugenol is added to `100 g` of camphor `(k_(f)=39.7" molality"^(-1))`, it lowered the freezing point by `0.62^(@)C` . The molar of eugenol is `:`A. `1.6xx10^(2) g//mol`B. `1.6xx10^(4)g//mol`C. `1.6xx10^(3)g//mol`D. `200g//mol` |
|
Answer» Correct Answer - 1 `0.62=(250xx10^(-3))/(Mxx100)xx1000xx39.7xx1` `M=160` or `1.6xx10^(2)g//mol` |
|