0.5 g of KCl was dissolved in 100 g of water and the solution originally at 20^@C froze at - 0.24^@C. Calculate the percentage dissociation of salt.Given : K_ffor water = 1.86 K kg/mol, At, mass : K= 39 u, Cl = 35.5 u.

0.5 g of KCl was dissolved in 100 g of water and the solution original

1.	0.5 g of KCl was dissolved in 100 g of water and the solution originally at 20^@C froze at - 0.24^@C. Calculate the percentage dissociation of salt.Given : K_ffor water = 1.86 K kg/mol, At, mass : K= 39 u, Cl = 35.5 u.
Answer» Solution :`Delta T_f = 0^@- (- 0.24^@C) = 0.24^@C "or" 0.24K` Apply the following equation ` M_2 = (K_f XX w_1 xx 1000)/((Delta T_f) xx w_1)` Substituting the values in this equation, we have `M_2 = (1.86 K kg "mol"^(-1) xx 0.5 g xx 1000 g kg^(-1))/(0.24 K xx 100 g) = 38.75 g "mol"^(-1)` van.t Hoff factor, i = Normal mol. mass / Abnormal mol. mass `= (74.5 g "mol"^(-1))/(38.75 g "mol"^(-1)) = 1.92` ` alpha = (i-1)/(n-1)` ,where n is the number of particles after dissociation In the case of KCL, n = 2 Substituting the values in the equation above `alpha = (1.92 - 1)/(2-1) " or " alpha = (0.92)/(1) = 0.91` Percentage dissociation = 0.92 x 100 = 92%