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_f` for water = 1.86 K kg/mol, At, mass : K= 39 u, Cl = 35.5 u.

Explanation

`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%