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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