#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)#
