One set Of recommendations for the salt to ice ratio for freezing homemade ice c

Question

One set Of recommendations for the salt to ice ratio for freezing homemade ice cream is to add 1.5 kg NaCl to 7.5 kg of ice. Assume half of the ice melts and % of the salt dissolves into that amount of melted water during the ice cream freezing process. a. What is the molality of the salt water solution? b. The freezing point depression constant for water is -1.86 degree C/m. Approximately how far would the salt solution lower the freezing point of the water? c. The expression for freezing point depression is based on the assumption of a dilute solution. Does this solution qualify as a dilute solution? And if not, would you expect characteristics of your solution to actually increase or decrease the change in the freezing point compared to the calculated value? Explain? (You should consider the mote ratio of water: NaCl in your explanation.)

Explanation / Answer

(7)

(a)

Mass of the ice melted = 7.5 / 2 = 3.75 kg.

Mass of NaCl dissolved = 1.5 * 3 / 4 = 1.125 kg. = 1125 g.

Moles of NaCl dissolved = mass / molar mass = 1125. / 58.5 = 19.2 mol

Molality of salt solution = moles of solute / weight of solvent in kg = 19.2 / 3.75 = 5.16 m

(b)

depression in freesing point = Kf * m

T0 - Tf = 1.86 * 5.16

Tf = 0.00 - 9.60 0C

Tf = Freezing point of solution = - 9.600C

(c)

moles of salt : moles of water = (3750/58.5) : 19.2 = 208.3 : 19.2

Compared to water, the amount of salt dissolved is too high, hence the solution is not dilute solution.

SO, the solution is highly concentrated and the depression in freezing point is directly proportional to molality of the solution. Hence more depression in freezing point can be observed than in dilute solutions.

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