

InterviewSolution
Saved Bookmarks
1. |
The density of `0.06 M` solution of `Kl` in water is `1.006 g mL^(-1)`. Determine the molality of this solution `(K = 39, I = 127 "amu")` |
Answer» `Mw of Kl = 39 + 127 = 166 g` `d = M ((Mw_(2))/(1000) + (1)/(m))` `1.006 = 0.006 ((166)/(1000) + (1)/(m))` Solve for `m`, `m = 0.06024` Alternate method: `M = (% "by Weight" xx 10 xx d)/(Mw_(2))` `0.06 = (%" "by Weight" xx 10 xx 1.006)/(166)` `:. "by weight" = (166 xx 0.06)/(10 xx 1.066) = 0.99%` Weight of solute `= 0.99 g`. Weight of solution `= 100 g` Weight of solvent `= 100 - 0.99 = 99.01` `m = (w_(2) xx 1000)/(Mw_(2) xx w_(1)) = (0.99 xx 1000)/(166 xx 99) = 0.06024` |
|