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)

Q!. 0.5 g of KCl was dissolved in 100 g of water and the solution orig

1.	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» 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 Kelvinchange in Tf = (i)(Kf)(m)i is van't hoff factorKf is freezing point depression constant.m is the molality.moles of solute= 0.5/(39 + 35.5)MASS of solvent in kg = 100/1000molality = moles of solute/ mass of solvent in kgKf 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.