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The freezing point of an aqueous solutions is 272 .93 K . Calculate the molality of the solution if molal depression constant for water is 1.86 Kg `mol^(-1)` |
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Answer» Correct Answer - Molality of the solution `=0.0376 m` Given : For pure water `T_(o) = 273 K` For solution `T_(r) = 273 .93 K` `K_(r) = 1.86 K kg mol^(-1)` Molality of the solution `= m= ?` Depression in the freezing point `= Delta T_(f)= T_(o) - T_(f)= 273 - 272 .93 = 0.07 K` `Delta T_(f) = K_(f) xx M` `:. m = (Delta T_(f))/(K_(f)) = (0.07)/(1.86) = 0.0376 "mol" kg ^(-1)` |
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