Question

Homemade ice-cream is frozen by churning it in a bucket suspended in an ice-water-salt mixture. A typical mix calls for 1.50 kg of salt (NaCl) and 8.25 kg of ice. Compute the mole fraction of NaCl in this mixture after all the ice melts.

0.0531

Compute the freezing point of this mixture (in °C).

Answer 1

Mass of NaCl in the mixture = 1.5 kg =1500 g

Molar mass of NaCl=58.44 g mol^-1

No. of moles of NaCl in 1500 g = 1500 g / (58.44 g mol^-1)

= 25.67 moles

Mass of ice in the mixture = 8.25 kg = 8250 g

Molar mass of water = 18 g mol^-1

No. of moles of water in 8250 g =  8250 g / (18 g mol^-1)

= 458.33 moles

Mole fraction of NaCl in the mixture = No. of moles NaCl / ( No. of moles NaCl + No. of moles water)

= 25.67 moles/ (25.67 moles + 458.33 moles)

= 0.053

The depression in freezing point in given by

$\Delta$T_f = i.K_f .m

$\Delta$T_f = depression in freezing point = T^o_f - T_f

T^o_f = Freezing point of pure water = 0^o C = 273.15 K

T_f = Freezing point of solution

i = Van't Hoff factor

As NaCl is a stong electrolyte, it will undergo complete dissociation.

Therefore, we will use i =2

K_f = Cryscopic constant of water = 1.853 K Kg mol^-1

m = molality of the solution

m= no. of moles of solute / mass of solvent in kg

= 25.67 mole / ( 8.25 kg)

= 3.112 moles / kg

Plugging the values in the equation ,

$\Delta$T_f = i.K_f .m

T^o_f - T_f = i.K_f .m

273.15 K - T_f = 2 * (1.853 K Kg mol^-1 )*[3.112 moles / kg ]

= 11.533 K

T_f = 273.15 K - 11.533 K

= 261.62 K

= (261.62 - 273.15 ) ^oC

= - 11.53 ^oC

Freezing point of mixture = - 11.53 ^oC