The freezing point of an aqueous sodium chloride solution is -0.20°C. What is the molality of the solution?

1 Answer
Jun 16, 2016

0.055\ (mol)/(Kg)

Explanation:

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)