Boiling and Freezing Points

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:08


We often encounter solutions in which the soluteThe substance added to a solvent to make a solution. has such a low vapor pressureThe pressure (or partial pressure) exerted by the gaseous form of a substance in equilibrium with the liquid form. as to be negligible. In such cases the vaporThe gaseous state of a substance that typically exists as a liquid or solid; a gas at a temperature near or below the boiling point of the corresponding liquid. above the solution consists only of solventThe substance to which a solute is added to make a solution. molecules and the vapor pressure is always lower than that of the pure solvent. Consider, for example, the solution obtained by dissolving 0.020 mol sucrose (C12H22O11) in 0.980 mol H2O at 100°C. The sucrose will contribute nothing to the vapor pressure, while we can expect the water vapor, by Raoult’s law, to contribute


PH2O = P*H2O × xH2O = 760 mmHg × 0.980 = 744.8 mmHg


We would thus expect the vapor pressure to be 744.8 mmHg, in reasonable agreement with the observed value of 743.3 mmHg.

A direct consequence of the lowering of the vapor pressure by a nonvolatile solute is an increase in the boiling pointThe temperature at which the vapor pressure of a liquid equals the pressure of the gas in contact with the liquid; usually this is atmospheric pressure. of the solution relative to that of the solvent. We can see why this is so by again using the sucrose solution as an example. At 100°C this solution has a vapor pressure which is lower than atmospheric pressure, and therefore it will not boil. In order to increase the vapor pressure from 743.3 to 760 mmHg so that boilingThe process of a liquid becoming vapor in which bubbles of vapor form beneath the surface of the liquid; at the boiling temperature the vapor pressure of the liquid equals the pressure of the gas in contact with the liquid. will occur, we need to raise the temperatureA physical property that indicates whether one object can transfer thermal energy to another object.. Experimentally we find that the temperature must be raised to 100.56°C. We say that the boiling-point elevation ΔTb is 0.56 K.

A second result of the lowering of the vapor pressure is a depression of the freezing pointThe temperature at which a liquid becomes a solid; also called melting point. of the solution. Any aqueousDescribing a solution in which the solvent is water. solution of a nonvolatile solute, for example, will have a vapor pressure at 0°C which is less than the vapor pressure of ice (0.006 atm or 4.6 mmHg) at this temperature. Accordingly, ice and the aqueous solution will not be in equilibriumA state in which no net change is occurring, that is, in which the concentrations of reactants and products remain constant; chemical equilibrium is characterized by forward and reverse reactions occurring at the same rate.. If the temperature is lowered, though, the vapor pressure of the ice decreases more rapidly than that of the solution and a temperature is soon reached when both ice and the solution have the same vapor pressure. Since both phases are now in equilibrium, this lower temperature is also the freezingThe process of forming a solid from a liquid. point of the solution. In the case of the sucrose solution of mole fractionIn a mixture, the chemical amount (moles) of one substance divided by the total chemical amount (moles) of all substances present. 0.02 described above, we find experimentally that the freezing point is –2.02°C.

We say that the freezing-point depression ΔTf is 2.02 K.

The depression of the freezing point of water by a solute explains why the sea does not freeze at 0°C. Because of its high saltAn ionic compound that can be formed by replacing the hydrogen ion of an acid with a different cation. content the sea has a freezing point of –2.2°C. If the sea froze at 0°C, larger stretches of ocean would turn into ice and the climate of the earth would be very different. We can now also understand why we add ethylene glycol, CH2OHCH2OH, to water in a car radiator in winter. Without any additive the water would freeze at 0°C and the resulting increase in volume would crack the radiator. Since ethylene glycol is very solubleAble to dissolve in a solvent to a significant extent. in water, it can form a solution with a freezing point low enough to prevent freezing even on the coldest winter day. Both the freezing-point depression and the boiling-point elevation of a solution were once important methods for determining the molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. of a newly prepared compoundA substance made up of two or more elements and having those elements present in definite proportions; a compound can be decomposed into two or more different substances.. Nowadays a mass spectrometer is usually used for this purpose, often on an assembly-line basis. Many chemists send samples of newly prepared compounds to a laboratory specializing in these determinations in much the same way as a medical doctor will send a sample of your blood to a laboratory for analysis.

The reason we can use the boiling-point elevation and the freezing-point depression to determine the molar mass of the solute is that both properties are proportional to the moleThat chemical amount of a substance containing the same number of units as 12 g of carbon-12. fraction and independent of the nature of the solute. The actual relationship, which we will not derive, is


x_{A}=\frac{\Delta H_{m}}{RT^{\text{2}}}\text{ }\Delta T


where xA is the mole fraction of the solute and ΔT is the boiling-point elevation or freezing-point depression. T indicates either the boiling point or freezing point of the pure solvent, and ΔHm is the molar enthalpyA thermodynamic state function, symbol H, that equals internal energy plus pressure x volume; the change in enthalpy corresponds to the energy transferred as a result of a temperature difference (heat transfer) when a reaction occurs at constant pressure. of vaporizationThe formation of a vapor from a liquid; evaporation or boiling. or fusionA nuclear reaction in which small nuclei are united to form larger ones., whichever is appropriate. This relationship tells us that we can measure the mole fraction of the solute in a solution merely by finding its boiling point or freezing point of the solution.



EXAMPLE 1 A solution of sucrose in water boils at 100.56°C and freezes at –2.02°C. Calculate the mole fraction of the solution from each temperature.


Solution

a) For boiling we have, from the Table of Molar Enthalpies of Fusion and Vaporization,


ΔHm = 40.7 kJ mol–1      and      T = 373.15 K


As in previous examples the units of R should be compatible with the other units appearing in the equation. In this case since ΔHm is given in units of kJ mol–1, R = 8.314 J K–1 mol–1 is most appropriate. Since ΔT = 0.56 K, we have


x_{\text{sucrose}}=\frac{\Delta H_{m}}{RT^{\text{2}}}\Delta \text{T}=\frac{\text{40}\text{.7 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}\text{ }\times \text{ 0}\text{.56 K}}{\text{8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ (373}\text{.15)}^{\text{2}}}=\text{0}\text{.0197}


b) Similarly for freezing we have


ΔHm = 6.01 kJ mol–1      T = 273.15 K      ΔT = 2.02 K


so that

x_{\text{sucrose}}=\frac{\text{6}\text{.01 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}\text{ }\times \text{ 2}\text{.02 K}}{\text{8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ }\times \text{ (273}\text{.15 K)}^{\text{2}}}=\text{0}\text{.0196}


The two values are in reasonable agreement.



Once we know the mole fraction of the solute, its molar mass is easily calculated from the mass composition of the solution.



EXAMPLE 2 33.07 g sucrose is dissolved in 85.27 g H2O. The resulting solution freezes at –2.02°C. Calculate the molar mass of sucrose.


Solution Since the freezing point of the solution is the same as in part b of Example 1, the mole fraction must be the same. Thus

                        x_{\text{sucrose}}=\text{0}\text{.0196}=\frac{n_{\text{sucrose}}}{n_{\text{sucrose}}\text{ + }n_{\text{H}_{\text{2}}\text{O}}}

Furthermore          \text{ }n_{\text{H}_{\text{2}}\text{O}}=\frac{\text{85}\text{.27 g}}{\text{18}\text{.02 g mol}^{-\text{1}}}=\text{4}\text{.732 mol}

Thus                 \text{0}\text{.0196}=\frac{n_{\text{sucrose}}}{n_{\text{sucrose}}\text{ + 4}\text{.732 mol}}


so that              \begin{align}
  & \text{4}\text{.732 }\times \text{ 0}\text{.0196 mol}=n_{\text{sucrose}}-\text{0}\text{.0196}n_{\text{sucrose}} \ 
 & \text{                                 0}\text{.092 75 mol}=\text{0}\text{.9804}n_{\text{sucrose}} \ 
\end{align}


or                     \text{ }n_{\text{sucrose}}=\frac{\text{0}\text{.092 75}}{\text{0}\text{.9804}}\text{mol}=\text{0}\text{.0946 mol}

From which      \text{  }M_{\text{sucrose}}=\frac{\text{33}\text{.07 g}}{\text{0}\text{.0946 mol}}=\text{350 g mol}^{-\text{1}}


Note: The correct molar mass is 342.3 g mol–1. Neither the freezing point nor the boiling point gives a very accurate value for the molar mass of the solute.