Q. 104.0( 1 Vote )

0.5 g of KCl was

Answer :


Change in Tf (freezing temperature) =


solvent's freezing point - solution's freezing point


Tf = T0f - Tf


On addition of solute, a freezing point always tends to decrease.


Hence , depression in freezing temperature= (0+273) - (0.240+273)


= -0.240K


Again depression in freezing point Tf = iKf m (i)


Where i is Van't Hoff factor and m is the molality of the solution and Kf is freezing point depression constant.


.


moles of the solute KCL =


mass of solvent water =


Now, molality =


Hence, normal molar mass of KCl = 74.5


Observed molar mass calculated from (i) is m = Kf X 1000


Tf X wsolvent


= 1.86 X .5 X 1000


.24 X 100


= 38.75


Vant hoff factor i = normal molar mass/observed molar mass


74.5/38.75 = 1.92


And


KCl dissociates as,


KCl K+ + Cl-


Moles after dissociation: 1 – α α α


Total no. Of moles after dissociation = 1 + α


hence , i = 1.92


or, α = 1.92 – 1 = 0.92


Hence, percentage dissociation of the salt = 92 %.


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