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Mole fraction of `I_(2)` in `C_(6) H_(6)` is 0.2. Calculate molality of `I_(2)` in `C_(6) H_(6)`. `(Mw of C_(6) H_(6) = 78 g mol^(-1))` |
Answer» `m = (x_(2) xx 1000)/(x_(1) xx Mw_(1)) = (0.2 xx 1000)/(0.8 xx 78) = 3.205` | |