0.5 g of KCl was dissolved in 100 g of water and the solution originally at 200° C, froze at -0.240C. Calculate the percentage dissociation of the salt. (Given :K_f for water = 1.86 K kg /mol, Atomic mass: K = 39 u, Cl= 35.5 u)

Question

Philoid · Accepted Answer

• Change in T_f (freezing temperature) =

solvent's freezing point - solution's freezing point

∆ T_f = T⁰_f - T_f

• 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 ∆ T_f = iK_f m (i)

Where i is Van't Hoff factor and m is the molality of the solution and K_f 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 = K_f X 1000

∆ T_f X w_solvent

= 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 %.

0.5 g of KCl was dissolved in 100 g of water and the solution originally at 200° C, froze at -0.240C. Calculate the percentage dissociation of the salt. (Given :Kf for water = 1.86 K kg /mol, Atomic mass: K = 39 u, Cl= 35.5 u)