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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))`A. `3.2`B. `6.40`C. `1.6`D. `2.30` |
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Answer» Correct Answer - A `(n)/(n+N) = 0.2 :. (N)/(n+N) = 0.8` or `(n)/(N) = (1)/(4)` or `(n_(I_(2)))/(w_(C_(6)H_(6))) xx M_(C_(6)H_(6)) = (1)/(4)` `:.` Molality `=(n_(I_(2)))/(w_(C_(6)H_(6))) xx 1000` `= (1)/(4) xx (1000)/(M_(C_(6)H_(6)))` or Molality `= (1)/(4) xx (1000)/(78)` `= 3.2` |
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