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22:3 Boiling Point and Freezing Point
22:3 BOILING POINT AND FREEZING POINT
The presence of nonvolatile solute particles at the surface causes the boiling point of a solution to be raised. The boiling point of a liquid is the aemperature at which the vapor pressure of the liquid equals the atmo- spheric pressure. In a solution, then, a higher temperature is needed to out enough solvent particles into the vapor phase to equal atmospheric pressure. The boiling point of a solution is, therefore, higher than that of the pure solvent. How does the addition of a nonvolatile solute affect the freezing Joint of a solution? The freezing point is the temperature at which the vapor pressure of the solid and liquid are equal. Since the addition of Solute particles lowers the vapor pressure, the vapor pressures of the solid and liquid will be equal at a lower temperature. Solutions, then, will freeze at a lower temperature than the pure solvent alone.
In summary, (^) the addition of a nonvolatile solute to a liquid causes both a boiling point elevation (^) and a freezing point depression. Both boi l - ing point elevation and freezing point depression occur because the vapor pressure of the solvent is lowered by the solute. These changes depend only on the concentration of the solute particles, and not upon the chem- ical composition of the solute. We will now consider some general quan- titative statements that can be made about these changes.
point
FIGURE 22-3. The addition of solute particles lowers the va- por pressure of a solvent and causes it to freeze at a lower temperature and boll at a tem- perature higher than normal.
point (^) Normal due to solute freezing point
424 Colligative and Colloidal Properties
22:4 CALCULATING FREEZING AND BOILING
POINTS
It has been found experimentally that 1 mole of nonvolatile solute particles will raise the boiling temperature of 1 kg of water 0.512 C°. The same concentration of solute will lower the freezing point of 1 kg of water 1.86 Co. These two figures are the molal boiling point constant and molal freezing point constant for water. The corresponding constants for some other common solvents are given in Table A-8 of the Appendix. A im solution of sugar in water contains 1 mol of solute particles per 1 kg of water. However, a 1 m solution of salt contains 2 mole of solute particles (1mol of Na} and 1 mol of Cl- ions). A 1m solution of calcium chloride contains 3 mol of solute particles per 1 kg of water (1 mol of Ca2+ and 2 mol of Cl- ions). The 1 m sugar solution freezes 1.86 C° below the freez- ing point of pure water. However, the 1m salt solution freezes about 2(1.86 C°) below the freezing point of pure water. The 1 m solution of calcium chloride freezes approximately 3(1.86 C°) below the freezing point of pure water. The multiple lowering of the freezing point and ele- vation of the boiling point by ionic substances supports the theory of dissociation.
EXAMPLE:^ Freezing Point Depression and
Boiling Point Elevation
If 85.0 grams of sugar are dissolved in 392 grams of water, what will be the boiling point and freezing point of the resulting solution? The molec- ular formula of sugar is C12H22011. Solving Process: (a) Determine the number of moles of solute. The mass of one mole of C12H22011 equals 342 g. Since there are 85.0 g of sugar present, the number of moles of sugar in 392 g of water is found by
85.03.C..^2111 mol C121122011 = 0.249 mol of C12H220n 342 g_C $ 1
(b) Convert this quantity to mol/1 000 g of water. The result is the molality of the solution. 0.249 mol C12H22011^ ,1000 9-water 392 ,g ..w^ ater I^1 kg water^
= (^) 0.635m solution
(c) Determine the boiling point elevation. The boiling point is raised 0.512^ C°^ for each mole of sugar added to 1000 g of water. Therefore, the boiling point is 100°C + (0.635m)(0.512 C°tm) = (100 + 0.325)°C = 100.325°C (d) Determine the freezing point depression. The freezing point is low-
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426 Colligative and Colloidal Properties
Compute the boiling and^ freezing^ points of the^ following^ solutions.
- 25.5 g C7H1 i NO7S^ (4-nitro-2-toluenesulfonic acid dehydrate)^ in 1. x 102 g H2O (nonionizing solute).
- 1.00 x 102 g C10H8O0S2 (1,5-naphthalenedisulfonic acid) in 1.00 x 102 g HzO (nonionizing solute)
- 21.6 g NiSO4 in 1.00 X 102 g H2O (assume 100% ionization)
- 77.0 g Mg(CIO4)2 in 2.00 x 102 g H2O (assume 100% ionization)
- 41.3 g C15H9N04^ (2-methyl-l-nitroanthraquinone)^ in 1.00 x 102 g C6H5N02^ (nitrobenzene,)
22:5 EXPERIMENTAL DETERMINATION OF
MOLECULAR MASS
Molecular mass of a solute may be determined by using boiling point elevation or freezing point depression. A known mass of the solute is added to known mass of a solvent. The resulting shift in the boiling or freezing point is then measured.
EXAMPLE: Molecular Mass Determination
99.0 grams of nonionizing solute are dissolved in 669 grams of water, and the freezing point of the resulting solution is -0.960°C.^ What is the molecular mass of the solute? Solving Process: (a) Determine the molality of the solution.
FIGURE 22-4. This drawing shows an apparatus used in molecular mass determinations based on freezing point depres- sion.
ATFP = (m)(KFP) m =
m=
(b) Calculate the molecular mass.
1 m = 0.516 mol/kg 860
ATFP KFP
Freezing tube^ 99.0 ,g solute 669,g water
molality
1000 g'^ I^ 1-kg 1_41 0.516 mol = - 287 gimol convert convert to kilograms^ tomoles
Calculate the^ molecular^ mass of the^ nonionic^ solutes.^ Use Table A-8 of the Appendix^ if necessary.
8. 8.02 grams of solute in 861 grams of water lower the freezing point to -0.430`C.
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22:6 Osmotic Pressure 427
- 64.3 grams of solute in 3.90 x 102 grams of water raise the boiling point to 100.680°C.
- 20.8 grams of^ solute in 128 grams of acetic acid lower the freezing point to 13.5CC.
- 10.4 grams^ of solute in 164 grams of phenol lower the freezing point to 36.3°C.
- 2.53 grams of solute in 63.5 grams of nitrobenzene lower the freez- ing point to 3.40°C.
22:6 OSMOTIC PRESSURE
There is another colligative property that is of great importance in wing systems. Consider Figure 22-5. Two liquids are separated by a thin film called a membrane. One liquid is a pure solvent and the other liquid s a solution of the same solvent. The membrane separating the liquids is a special kind of membrane called a semipermeable membrane. Semiper- meable membranes will allow small particles (ions and molecules) to pass through, but will stop large molecules. As a result of an unequal passing of particles, a pressure difference builds up between the two sides of the membrane. This pressure is called the osmotic pressure of the solution. Osmotic processes are very important in the human body. Absorption of the products of digestion and the operation of the kidney as a waste remover are two osmotic processes vital to life.
Since osmotic pressure is a colligative property, it can be expressed in an equation such as TI = (m)(Kos,,,)
where TI is the osmotic pressure. However,^ the constant^ for osmosis
FIGURE 22 - 5. Osmotic pressure can be measured as the force applied to the piston to oppose the osmotic flow through the membrane.
