Delta T= K_fxxmxx i
where:
m -> " is the molality of the solute expressed in " (mol)/(Kg)
Delta T ->" is the difference between the freezing point of the "
"solution"\ (T_f)" and that of the pure solvent "(T_i) .
\Delta T = T_f- T_i
i -> " is the Van'to Hoff factor. It is related to the number of"
"particles ( or ions) of the solute dissolved in the solution."
"in the case of " NaCl " " i = 2 " since " NaCl" fully ionizes into 2 ions"
K_f->"is the freezing point depression constant of the solvent."
"in aqueous solutions the solvent is water and " K_f = 1.86\°C*Kg*mol^-1
Delta T= - K_f\ xxunderbrace (mxxi)_("molality of particles")
Delta T= - K_f\ xxunderbrace (mxxi)_(m')
Delta T= - K_fxx\ m'
m'=- (Delta T)/ K_f
m'=- ((T_f-T_i))/ K_f
m'= -((-0.20 °C-0°C))/(1.86°C.Kg.mol^-1)
m'= (0.20 \ cancel (°C))/(1.86cancel (°C.)Kg.mol^-1)
m'= 0.11\ (mol)/(Kg)
Note that m' is the molality of the dissolved particles in the solution i.e \ Na^+ + Cl^-1 while m is the molality of the solute (NaCl)
mxxi=m'
m = 0.11 /2\ (mol)/(Kg)
m =0.055 (mol)/(Kg)