Question

Q!. 0.5 g of KCl was dissolved in 100 g of water and the solution originally at 200
C, froze at
-0.240
C. 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)

Answer 1

change in Tf (freezing temperature) = solvent's freezing point - solution's freezing point. (coz on addition of solute, freezing point tends to decrease)
change in Tf = (0+273) - (-0.240+273) = 0.240 Kelvin
change in Tf = (i)(Kf)(m)
i is van't hoff factor
Kf is freezing point depression constant.
m is the molality.
moles of solute= 0.5/(39 + 35.5)
mass of solvent in kg = 100/1000
molality = moles of solute/ mass of solvent in kg
Kf is given.
solve for i.
next, (alpha) = (i-1)/(n-1)
n is 2, since no. of moles formed after dissociation of kcl in water is 2 (k+ and cl-)
alpha/100 is percentage dissociation.