Video: Colligative Properties: Freezing Point Depression

Which of the following aqueous solutions has the lowest freezing point? [A] 0.080 m Na₃PO₄ [B] 0.10 m C₂H₅OH [C] 0.010 m CO₂ [D] 0.15 m CaCL₂ [E] 0.12 m NaCL.

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Video Transcript

Which of the following aqueous solutions has the lowest freezing point? a) 0.080 molal Na₃PO₄, b) 0.10 molal C₂H₅OH, c) 0.010 molal CO₂, d) 0.15 molal CaCL₂, or e) 0.12 molal NaCL.

For each of these solutions, the solvent is water. The concentrations of these solutions are given in molal rather molar. This is equivalent to moles per kilogram rather than moles per liter. So the molal concentration is equal to the amount of solute in moles divided by the mass of solvent in kilograms. So what’s all this got to do with the lowest freezing point solution? Well, the freezing point of pure water is zero degrees Celsius. The freezing point of a solution will be slightly lower. This decrease in freezing point is sometimes called the freezing point depression. And it can be calculated for any solution by combining the van ’t Hoff factor, the cryoscopic constant for the solvent, and the solution’s molality.

The cryoscopic constant is unique to each solvent in the case of water is equal to 0.512 kelvins kilogram per mole. The van ’t Hoff factor is equal to the number of particles the solute forms. In the case of a salt like sodium chloride, the van ’t Hoff factor is equal to two because one unit of sodium chloride dissolves to form two separate ions. Before we go any further, we should figure out the van ’t Hoff factor for each of our substances.

Sodium phosphate dissolves easily to form three sodium ions and one phosphate ion. So it has a van ’t Hoff factor of four. Ethanol C₂H₅OH dissolves as a single molecule. So it has van ’t Hoff factor of one. The van ’t Hoff factor of carbon dioxide is also one. Carbon dioxide does dissolve in water to form carbonic acid. But for the purposes of this question, we can assume that it forms just a single molecule. Calcium chloride dissolves to form a calcium ion and two chloride ions. So it has a van ’t Hoff factor of three.

Now, we could actually calculate the value of the freezing point depression. But since there are only comparing solutions and we want to find the one with the lowest freezing point, all we need to do is calculate the Δ𝑇 in terms of 𝐾 𝑏. So for our first solution, Δ𝑇 is equal to four times 𝐾 𝑏 times 0.080, giving us a value of 0.32 𝐾 𝑏. We’re ignoring units of molal for the moment. Δ𝑇 for our 0.10 molal ethanol solution is one times 𝐾 𝑏 times 0.10, giving us 0.10 𝐾 𝑏. For our carbon dioxide solution, we get a value of 0.010 𝐾 𝑏. For our calcium chloride solution, we get 0.45. And for our sodium chloride solution, we get 0.24. So of all of our values, the highest one is 0.45 𝐾 𝑏, giving us the greatest Δ𝑇 and the lowest freezing point for our calcium chloride solution.

If you like, we can work out the actual freezing point depression for this solution. It’s equal to three times 0.512 kelvins kilograms per mole multiplied by 0.15 moles per kilogram, giving us a value of 0.2304 Kelvin for the freezing point depression and freezing point of negative 0.23 degrees Celsius to two significant figures. Just be careful to add in the units for the molals if you want to work out any of the other values. Of the five aqueous solutions given, the one with the lowest freezing point at negative 0.23 degrees Celsius is our 0.15 molal calcium chloride solution.

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