Have you ever wondered what they are sparing on airplanes on cold days before they take off?  What they are spraying is a solution typically containing ethylene glycol, C₂H₆O₂ which lowers the freezing point of water. It is essentially the same solution that is used as antifreeze in cars. We can also see this in the phase diagram of the solvent and the solution. It includes two regions of interest for the colligative properties. On the left side, is the freezing/melting point region, and on the right side, toward higher temperatures, we have the boiling point region. We have already discussed the boiling point elevation of the solution, so let’s now see how the freezing point depression is explained. On the x axis, we have the temperature and as it increases, the phase is transitioning from solid-liquid-gas. Notice, on the left side, that the freezing point of the solution is lower than that of the solvent. To understand this property, let’s again recall that the entropy of a solution is higher than the entropy of the solvent because the mixing of components increases the randomness of the system. Freezing, on the other hand, decreases the randomness, and thus the process requires energy to be removed from the system to create a more ordered state. Now, because the solution has higher entropy, it requires more energy to be removed to transition to a solid phase. Therefore, a solution has a lower freezing point than the solvent. The Freezing point depression is calculated with the following formula: Where Kf is the molal freezing point depression constant characteristic of a given solvent. The values of Kf for some common solvents are given below: and m is the molality of the solute which is calculated using this formula: For example, using the appropriate data in the table, determine the freezing point depression of the solution that contains 24.1 g urea (NH2)2CO) in 485 mL of water. The freezing point depression is calculated by the following formula: ΔTf = m x Kf Kf is the freezing point depression constant for the solvent, and from the table, we find that it is 1.86 oC/m m is the molality of the solute which is equal to: \[{\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{mol}}\;{\rm{(solute)}}}}{{{\rm{kg}}\,{\rm{(solvent)}}}}\] The moles of urea are calculated from the mass: \[{\rm{n}}\,{\left( {{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)_{\rm{2}}}{\rm{CO)}}\;{\rm{ = }}\,{\rm{24}}{\rm{.1}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{60}}{\rm{.1}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.401}}\;{\rm{mol}}\] Convert the mL to kg for water: \[{\rm{m}}\,{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{O)}}\;{\rm{ = }}\,{\rm{485}}\,\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}{\rm{.00}}\,\cancel{{\rm{g}}}}}{{{\rm{1}}\,\cancel{{{\rm{mL}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{kg}}}}{{{\rm{1000}}\;\cancel{{\rm{g}}}}}{\rm{ = }}\,{\rm{0}}{\rm{.485}}\,{\rm{kg}}\] The molality of the solute is: \[{\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.401}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.485}}\,{\rm{kg}}}}\; = \;0.827\,m\] The freezing point depression would then be: ΔTf = Kfm = 1.86 oC /m x 0.827 m = 0.902 oC As in the case of vapor pressure lowering, when a strong electrolyte is dissolved in water, the concentration of the solute particles is given by the ions rather than the formula of the compound because strong electrolytes dissociate into ions in aqueous solutions. So, if we dissolve 1 mole of NaCl in water it will dissociate into ions and two moles of ions will be formed: NaCl(aq) → Na+(aq) + Cl–(aq) In general, we can calculate the number of ions based on the formula of the salt. For example, 1 mole of MgBr2 is expected to produce 3 moles of ions because each formula unit contains one Mg2+ and 2 Br– ions. However, the dissociation of most ionic compounds does not occur at 100%, and the solution of an ionic compound usually contains fewer particles than what its formula suggests. The actual extent of dissociation can be expressed as a van’t Hoff factor (i). For most ionic compounds, the van’t Hoff’s constant is determined experimentally, and will likely be given to you in the test. If it is not, and there is nothing mentioned about, you can go based on the formula of the compound. For nonelectrolytes, it is assumed to be 1 as we do not worry about their negligible dissociation. For example, calculate the freezing point of the solution prepared by dissolving 4.80 g NaCl in 25.0 g of water. The freezing point depression is calculated by the following formula: ΔTf = m x Kf Kf is the freezing point depression constant for the solvent, and from the table, we find that it is 1.86 oC/m m is the molality of the solute which is equal to: \[{\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{mol}}\;{\rm{(solute)}}}}{{{\rm{kg}}\,{\rm{(solvent)}}}}\] The moles of particles from 4.80 g NaCl are calculated by multiplying the moles of NaCl by two since it dissociates into two ions: \[{\rm{n}}\,({\rm{particles)}}\;{\rm{ = }}\,2\,{\rm{ \times }}\;{\rm{4}}{\rm{.80}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{58}}{\rm{.44}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.164}}\;{\rm{mol}}\] The mass of water is 0.0250 kg, therefore, the molality of the solute is: \[{\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.164}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0250}}\,{\rm{kg}}}}\; = \;6.56\,m\] The freezing point depression would then be: ΔTf = Kfm = 1.86 oC /m x 6.56 m = 12.2 oC The normal freezing point of water is 0oC, therefore, the freeing point of the solution will be: 0 – 12.2 = -12.2 oC For more practice problems on the colligative properties, follow this link. 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In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. Colligative properties have practical applications, such as the salting of roads in cold-weather climates. By applying salt to an icy road, the melting point of the ice is decreased, and the ice will melt more quickly, making driving safer. Sodium chloride \(\left( \ce{NaCl} \right)\), and either calcium chloride \(\left( \ce{CaCl_2} \right)\) or magnesium chloride \(\left( \ce{MgCl_2} \right)\) are used most frequently, either alone or in a mixture. Sodium chloride is the least expensive option, but is less effective because it only dissociates into two ions, instead of three. The figure below shows the phase diagram for a pure solvent and how it changes when a solute is added to it. The solute lowers the vapor pressure of the solvent, resulting in a lowering of the freezing point of the solution compared to the solvent. The freezing point depression is the difference in temperature between the freezing point of the pure solvent and that of the solution. On the graph, the freezing point depression is represented by \(\Delta T_f\). When a pure solvent freezes, its particles become more ordered as the intermolecular forces that operate between the molecules become permanent. In the case of water, the hydrogen bonds make the hexagonally-shaped network of molecules that characterizes the structure of ice. By dissolving a solute into the liquid solvent, this ordering process is disrupted. As a result, more energy must be removed from the solution in order to freeze it, and the freezing point of the solution is lower than that of the pure solvent. The magnitude of the freezing point depression is directly proportional to the molality of the solution. The equation is: \[\Delta T_f = K_f \times \textit{m}\nonumber \] The proportionality constant, \(K_f\), is called the molal freezing-point depression constant. It is a constant that is equal to the change in the freezing point for a 1-molal solution of a nonvolatile molecular solute. For water, the value of \(K_f\) is \(-1.86^\text{o} \text{C}/\textit{m}\). So, the freezing temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is \(-1.86^\text{o} \text{C}\). Every solvent has a unique molal freezing-point depression constant. These are shown in the table below, along with a related value for the boiling point called \(K_b\).
Ethylene glycol \(\left( \ce{C_2H_6O_2} \right)\) is a molecular compound that is used in many commercial antifreezes. A water solution of ethylene glycol is used in vehicle radiators to lower its freezing point, and thus prevent the water in the radiator from freezing. Calculate the freezing point of a solution of \(400 \: \text{g}\) of ethylene glycol in \(500 \: \text{g}\) of water. This is a three-step problem. First, calculate the moles of ethylene glycol. Then, calculate the molality of the solution. Finally, calculate the freezing point depression.
\[\begin{align*} 400. \: \text{g} \: \ce{C_2H_6O_2} \times \frac{1 \: \text{mol} \: \ce{C_2H_6O_2}}{62.08 \: \text{g} \: \ce{C_2H_6O_2}} &= 6.44 \: \text{mol} \: \ce{C_2H_6O_2} \\ \frac{6.44 \: \text{mol} \: \ce{C_2H_6O_2}}{0.500 \: \text{kg} \: \ce{H_2O}} &= 12.9 \: \textit{m} \: \ce{C_2H_6O_2} \\ \Delta T_f = K_f \times \textit{m} = -1.86^\text{o} \text{C}/\textit{m} &= -24.0^\text{o} \text{C} \\ T_f &= -24.0^\text{o} \text{C} \end{align*}\nonumber \] The normal freezing point of water is \(0.0^\text{o} \text{C}\). Therefore, since the freezing point decreases by \(24.0^\text{o} \text{C}\), the freezing point of the solution is \(-24.0^\text{o} \text{C}\).
The freezing point of the water decreases by a large amount, protecting the radiator from damage due to the expansion of water when it freezes. There are three significant figures in the result. Summary
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Freezing point depression constant of ethylene glycol
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