Question

A mixture of table salt and ice is used to chill the contents of hand-operated ice cream makers. What is the melting point of a mixture of 2.00 lb of NaCl and 12.00 pounds of ice if exactly half of the ice melts? Assume that all the NaCl disolves in the melted ice.

Answer 1

∆Tf =i x Kf x m

i = Van't Hoff Factor (number of particles dissociatied either experimentally or theoretically)

Kf = Freezing point depression constant of solvent

m = molality of solvent

The experimental value of i = 1.44 .

Kf value of water = 1.86°C/molal.

2 lbs. NaCl = 907.185 g NaCl

6 lbs. = 2.722 kg Water

moles of NaCl = mass / moalr mass = 907.185 / 58.5

= 15.5 mol NaCl

molality = moles of solute / mass of solvent in kg

= 15.5 / 2.722kg

= 5.7 molal

Solve for the change in freezing point:

∆Tf= (1.44)(1.86°C/m)(5.7m) = 15.27°C

To - Tf = 15.27 ( The normal freezing point of water is 0°C,)

0 - Tf = 15.27

Tf = - 15.27

The resulting freezing temperature is -15.27°C.