InterviewSolution
Saved Bookmarks
| 1. |
When `250 mg` of eugenol is added to `100 g` of camphor `(K_(f) = 39.7K "molality"^(-1))`, it lowered the freezing point by `0.62^(@)C`. The molar mass of eugenol is:A. (a) `160g mol^(-1)`B. (b) `165g mol^(-1)`C. (c ) `200g mol^(-1)`D. (d) `250g mol^(-1)` |
|
Answer» Correct Answer - A `m = (1000 xx K_(f) xx w)/(W xx Delta T)` `= (1000 xx 39.7 xx 250 xx 10^(-3))/(100 xx 0.62)` `= 160 g mol^(-1)` |
|