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

Answer:

#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)#