Question

0.5 gm of Kcl was dissolved in 100 gm of water 20°c - 0.24°c calculate percent insoniation of salt kf= 1.86​

Answer 1

• 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 %.

Answer 2

Answer:

92% is the correct answer

Explanation:

92% is the correct answer