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The freezing point of an aqueous sol...

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)`

Text Solution

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The correct Answer is:
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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