{"id":7584,"date":"2021-06-08T21:56:55","date_gmt":"2021-06-08T21:56:55","guid":{"rendered":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/colligative-properties-of-solutions\/"},"modified":"2021-10-06T20:14:13","modified_gmt":"2021-10-06T20:14:13","slug":"colligative-properties-of-solutions","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/introductorychemistry\/chapter\/colligative-properties-of-solutions\/","title":{"raw":"Colligative Properties of Solutions","rendered":"Colligative Properties of Solutions"},"content":{"raw":"[latexpage]\r\n
\r\n

Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

\r\n
    \r\n \t
  1. Name the four colligative properties.<\/li>\r\n \t
  2. Calculate changes in vapour pressure, melting point, and boiling point of solutions.<\/li>\r\n \t
  3. Calculate the osmotic pressure of solutions.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\nThe properties of solutions are very similar to the properties of their respective pure solvents. This makes sense because the majority of the solution is<\/em> the solvent. However, some of the properties of solutions differ from pure solvents in measurable and predictable ways. The differences are proportional to the fraction that the solute particles occupy in the solution. These properties are called colligative properties<\/strong>; the word colligative<\/em> comes from the Greek word meaning \u201crelated to the number,\u201d implying that these properties are related to the number of solute particles, not their identities.\r\n\r\nBefore we introduce the first colligative property, we need to introduce a new concentration unit. The mole fraction<\/strong>\u00a0of the i<\/em>th component in a solution, \u03c7i<\/em><\/sub>, is the number of moles of that component divided by the total number of moles in the sample:\r\n

    [latex]\\chi_i=\\dfrac{\\text{ moles of }i\\text{th component}}{\\text{total moles}}[\/latex]<\/p>\r\n(\u03c7 is the lowercase Greek letter chi.) The mole fraction is always a number between 0 and 1 (inclusive) and has no units; it is just a number.\r\n

    \r\n

    Example 11.1<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nA solution is made by mixing 12.0 g of C10<\/sub>H8<\/sub> in 45.0 g of C6<\/sub>H6<\/sub>. What is the mole fraction of C10<\/sub>H8<\/sub> in the solution?\r\n\r\nSolution<\/em>\r\n\r\nWe need to determine the number of moles of each substance, add them together to get the total number of moles, and then divide to determine the mole fraction of C10<\/sub>H8<\/sub>. The number of moles of C10<\/sub>H8<\/sub> is as follows:\r\n

    [latex]12.0\\cancel{\\text{ g }\\ce{C10H8}}\\times\\dfrac{1\\text{ mol }\\ce{C10H8}}{128.18\\cancel{\\text{ g }\\ce{C10H8}}}=0.0936\\text{ mol }\\ce{C10H8}[\/latex]<\/p>\r\nThe number of moles of C6<\/sub>H6<\/sub> is as follows:\r\n

    [latex]45.0\\cancel{\\text{ g }\\ce{C6H6}}\\times\\dfrac{1\\text{ mol }\\ce{C6H6}}{78.12\\cancel{\\text{ g }\\ce{C6H6}}}=0.576\\text{ mol }\\ce{C6H6}[\/latex]<\/p>\r\nThe total number of moles is:\r\n

    [latex]0.0936\\text{ mol}+ 0.576\\text{ mol}= 0.670\\text{ mol}[\/latex]<\/p>\r\nNow we can calculate the mole fraction of C10<\/sub>H8<\/sub>:\r\n

    [latex]\\chi_{\\ce{C10H8}}=\\dfrac{0.0936\\text{ mol}}{0.670\\text{ mol}}=0.140[\/latex]<\/p>\r\nThe mole fraction is a number between 0 and 1 and is unitless.\r\n\r\nTest Yourself<\/em>\r\n\r\nA solution is made by mixing 33.8 g of CH3<\/sub>OH in 50.0 g of H2<\/sub>O. What is the mole fraction of CH3<\/sub>OH in the solution?\r\n\r\nAnswer<\/em>\r\n\r\n0.275\r\n\r\n<\/div>\r\n<\/div>\r\nA useful thing to note is that the sum of the mole fractions of all substances in a mixture equals 1. Thus the mole fraction of C6<\/sub>H6<\/sub> in Example 11.1 could be calculated by evaluating the definition of mole fraction a second time, or \u2014 because there are only two substances in this particular mixture \u2014 we can subtract the mole fraction of the C10<\/sub>H8<\/sub> from 1 to get the mole fraction of C6<\/sub>H6<\/sub>.\r\n\r\nNow that this new concentration unit has been introduced, the first colligative property can be considered. As was mentioned in Chapter 10 \"Solids and Liquids\"<\/a>, all pure liquids have a characteristic vapour pressure in equilibrium with the liquid phase, the partial pressure of which is dependent on temperature. Solutions, however, have a lower vapour pressure than the pure solvent has, and the amount of lowering is dependent on the fraction of solute particles, as long as the solute itself does not have a significant vapour pressure (the term nonvolatile<\/em> is used to describe such solutes). This colligative property is called vapour pressure depression<\/strong>\u00a0(or lowering<\/em>). The actual vapour pressure of the solution can be calculated as follows:\r\n

    [latex]P_{\\text{soln}}=\\chi_{\\text{solv}}P^\\ast_{\\text{solv}}[\/latex]<\/p>\r\nwhere P<\/em>soln<\/sub> is the vapour pressure of the solution, \u03c7solv<\/sub> is the mole fraction of the solvent particles, and P*<\/sup>solv<\/sub> is the vapour pressure of the pure solvent at that temperature (which is data that must be provided). This equation is known as Raoult\u2019s law<\/strong>\u00a0(the approximate pronunciation is rah-OOLT<\/em>). Vapour pressure depression is rationalized by presuming that solute particles take positions at the surface in place of solvent particles, so not as many solvent particles can evaporate.\r\n

    \r\n

    Example 11.2<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nA solution is made by mixing 12.0 g of C10<\/sub>H8<\/sub> in 45.0 g of C6<\/sub>H6<\/sub>. If the vapour pressure of pure C6<\/sub>H6<\/sub> is 95.3 torr, what is the vapour pressure of the solution?\r\n\r\nSolution<\/em>\r\n\r\nThis is the same solution that was in Example 11.1, but here we need the mole fraction of C6<\/sub>H6<\/sub>. The number of moles of C10<\/sub>H8<\/sub> is as follows:\r\n

    [latex]12.0\\cancel{\\text{ g }\\ce{C10H8}}\\times\\dfrac{1\\text{ mol }\\ce{C10H8}}{128.18\\cancel{\\text{ g }\\ce{C10H8}}}=0.0936\\text{ mol }\\ce{C10H8}[\/latex]<\/p>\r\nThe number of moles of C6<\/sub>H6<\/sub> is as follows:\r\n

    [latex]45.0\\cancel{\\text{ g }\\ce{C6H6}}\\times\\dfrac{1\\text{ mol }\\ce{C6H6}}{78.12\\cancel{\\text{ g }\\ce{C6H6}}}=0.576\\text{ mol }\\ce{C6H6}[\/latex]<\/p>\r\nSo the total number of moles is:\r\n

    [latex]0.0936\\text{ mol}+0.576\\text{ mol}=0.670\\text{ mol}[\/latex]<\/p>\r\nNow we can calculate the mole fraction of C6<\/sub>H6<\/sub>:\r\n

    [latex]\\chi_{\\ce{C6H6}}=\\dfrac{0.576\\text{ mol}}{0.670\\text{ mol}}=0.860[\/latex]<\/p>\r\n(The mole fraction of C10<\/sub>H8<\/sub> calculated in Example 11.1 plus the mole fraction of C6<\/sub>H6<\/sub> equals 1, which is mathematically required by the definition of mole fraction.) Now we can use Raoult\u2019s law to determine the vapour pressure in equilibrium with the solution:\r\n

    [latex]P_{\\text{soln}}=(0.860)(95.3\\text{ torr})=82.0\\text{ torr}[\/latex]<\/p>\r\nThe solution has a lower vapour pressure than the pure solvent.\r\n\r\nTest Yourself<\/em>\r\n\r\nA solution is made by mixing 33.8 g of C6<\/sub>H12<\/sub>O6<\/sub> in 50.0 g of H2<\/sub>O. If the vapour pressure of pure water is 25.7 torr, what is the vapour pressure of the solution?\r\n\r\nAnswer<\/em>\r\n24.1 torr\r\n\r\n<\/div>\r\n<\/div>\r\nTwo colligative properties are related to solution concentration as expressed in molality. As a review, recall the definition of molality:\r\n

    [latex]\\text{molality}=\\dfrac{\\text{moles solute}}{\\text{kilograms solvent}}[\/latex]<\/p>\r\nBecause the vapour pressure of a solution with a nonvolatile solute is depressed compared to that of the pure solvent, it requires a higher temperature for the solution\u2019s vapour pressure to reach 1.00 atm (760 torr). Recall that this is the definition of the normal boiling point: the temperature at which the vapour pressure of the liquid equals 1.00 atm. As such, the normal boiling point of the solution is higher than that of the pure solvent. This property is called boiling point elevation<\/strong>.\r\n\r\nThe change in boiling point (\u0394T<\/em>b<\/sub>) is easily calculated:\r\n

    [latex]\\Delta T_\\text{b}=mK_\\text{b}[\/latex]<\/p>\r\nwhere m<\/em> is the molality of the solution and K<\/em>b<\/sub> is called the boiling point elevation constant<\/strong>, which is a characteristic of the solvent. Several boiling point elevation constants (as well as boiling point temperatures) are listed in Table 11.2 \"Boiling Point Data for Various Liquids\".\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Table 11.2 Boiling Point Data for Various Liquids<\/caption>\r\n
    Liquid<\/th>\r\nBoiling Point (\u00b0C)<\/th>\r\nK<\/em>b<\/sub> (\u00b0C\/m<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    HC2<\/sub>H3<\/sub>O2<\/sub><\/td>\r\n117.90<\/td>\r\n3.07<\/td>\r\n<\/tr>\r\n
    C6<\/sub>H6<\/sub><\/td>\r\n80.10<\/td>\r\n2.53<\/td>\r\n<\/tr>\r\n
    CCl4<\/sub><\/td>\r\n76.8<\/td>\r\n4.95<\/td>\r\n<\/tr>\r\n
    H2<\/sub>O<\/td>\r\n100.00<\/td>\r\n0.512<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRemember that what is initially calculated is the change<\/em> in boiling point temperature, not the new boiling point temperature. Once the change in boiling point temperature is calculated, it must be added to the boiling point of the pure solvent \u2014 because boiling points are always elevated \u2014 to get the boiling point of the solution.\r\n
    \r\n

    Example 11.3<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is the boiling point of a 2.50 m<\/em> solution of C6<\/sub>H4<\/sub>Cl2<\/sub> in CCl4<\/sub>? Assume that C6<\/sub>H4<\/sub>Cl2<\/sub> is not volatile.\r\n\r\nSolution<\/em>\r\n\r\nUsing the equation for the boiling point elevation:\r\n

    [latex]\\Delta T_{\\text{b}}=(2.50 \\cancel{m})\\left(\\dfrac{4.95^{\\circ}\\text{ C}}{\\cancel{m}}\\right)=12.4^{\\circ}{\\text{ C}}[\/latex]<\/p>\r\nNote how the molality units have cancelled. However, we are not finished. We have calculated the change in the boiling point temperature, not the final boiling point temperature. If the boiling point goes up by 12.4\u00b0C, we need to add this to the normal boiling point of CCl4<\/sub> to get the new boiling point of the solution:\r\n

    [latex]T_{\\text{BP}}=76.8^{\\circ}\\text{C}+12.4^{\\circ}\\text{C}=89.2^{\\circ}\\text{C}[\/latex]<\/p>\r\nThe boiling point of the solution is predicted to be 89.2\u00b0C.\r\n\r\nTest Yourself<\/em>\r\n\r\nWhat is the boiling point of a 6.95 m<\/em> solution of C12<\/sub>H22<\/sub>O11<\/sub> in H2<\/sub>O?\r\n\r\nAnswer<\/em>\r\n103.6\u00b0C\r\n\r\n<\/div>\r\n<\/div>\r\nThe boiling point of a solution is higher than the boiling point of the pure solvent, but the opposite occurs with the freezing point. The freezing point of a solution is lower than the freezing point of the pure solvent. Think of this by assuming that solute particles interfere with solvent particles coming together to make a solid, so it takes a lower temperature to get the solvent particles to solidify. This is called freezing point depression<\/strong>.\r\n\r\nThe equation to calculate the change in the freezing point for a solution is similar to the equation for the boiling point elevation:\r\n

    [latex]\\Delta T_{\\text{f}}=mK_{\\text{f}}[\/latex]<\/p>\r\nwhere m<\/em> is the molality of the solution and K<\/em>f<\/sub> is called the freezing point depression constant<\/strong>, which is also a characteristic of the solvent. Several freezing point depression constants (as well as freezing point temperatures) are listed in Table 11.3 \"Freezing Point Data for Various Liquids\".\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Table 11.3<\/span> Freezing Point Data for Various Liquids<\/caption>\r\n
    Liquid<\/th>\r\nFreezing Point (\u00b0C)<\/th>\r\nK<\/em>f<\/sub> (\u00b0C\/m<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    HC2<\/sub>H3<\/sub>O2<\/sub><\/td>\r\n16.60<\/td>\r\n3.90<\/td>\r\n<\/tr>\r\n
    C6<\/sub>H6<\/sub><\/td>\r\n5.51<\/td>\r\n4.90<\/td>\r\n<\/tr>\r\n
    C6<\/sub>H12<\/sub><\/td>\r\n6.4<\/td>\r\n20.2<\/td>\r\n<\/tr>\r\n
    C10<\/sub>H8<\/sub><\/td>\r\n80.2<\/td>\r\n6.8<\/td>\r\n<\/tr>\r\n
    H2<\/sub>O<\/td>\r\n0.00<\/td>\r\n1.86<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRemember that this equation calculates the change in the freezing point, not the new freezing point. What is calculated needs to be subtracted from the normal freezing point of the solvent because freezing points always go down.\r\n
    \r\n

    Example 11.4<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is the freezing point of a 1.77 m<\/em> solution of CBr4<\/sub> in C6<\/sub>H6<\/sub>?\r\n\r\nSolution<\/em>\r\n\r\nWe use the equation to calculate the change in the freezing point and then subtract this number from the normal freezing point of C6<\/sub>H6<\/sub> to get the freezing point of the solution:\r\n

    [latex]\\Delta T_{\\text{f}}=(1.77 \\cancel{m})\\left(\\dfrac{4.90^{\\circ}\\text{C}}{\\cancel{m}}\\right)=8.67^{\\circ}\\text{C}[\/latex]<\/p>\r\nNow we subtract this number from the normal freezing point of C6<\/sub>H6<\/sub>, which is 5.51\u00b0C:\r\n

    [latex]5.51^{\\circ}\\text{C}-8.67^{\\circ}\\text{C}=-3.16^{\\circ}\\text{C}[\/latex]<\/p>\r\nThe freezing point of the solution is \u22123.16\u00b0C.\r\n\r\nTest Yourself<\/em>\r\n\r\nWhat is the freezing point of a 3.05 m<\/em> solution of CBr4<\/sub> in C10<\/sub>H8<\/sub>?\r\n\r\nAnswer<\/em>\r\n59.5\u00b0C\r\n\r\n<\/div>\r\n<\/div>\r\nFreezing point depression is one colligative property we use in everyday life. Many antifreezes used in automobile radiators use solutions that have a lower freezing point than normal so that automobile engines can operate at subfreezing temperatures. We also take advantage of freezing point depression when we sprinkle various compounds on ice to thaw it in the winter for safety (see Figure 11.2 \"Salt and Safety\"). The compounds make solutions that have a lower freezing point, so rather than forming slippery ice, any ice is liquefied and runs off, leaving a safer pavement behind.\r\n\r\n[caption id=\"attachment_625\" align=\"aligncenter\" width=\"324\"]\"Salt Figure 11.2 \"Salt and Safety.\" Salt or other compounds take advantage of the freezing point depression to minimize the formation of ice on sidewalks and roads, thus increasing safety.[\/caption]\r\n\r\nBefore we introduce the final colligative property, we need to present a new concept. A semipermeable membrane<\/strong>\u00a0is a thin membrane that will pass certain small molecules but not others. A thin sheet of cellophane, for example, acts as a semipermeable membrane.\r\n\r\nConsider the system in part (a) of Figure 11.3 \"Osmosis.\" A semipermeable membrane separates two solutions having the different concentrations marked. Curiously, this situation is not stable; there is a tendency for water molecules to move from the dilute side (on the left) to the concentrated side (on the right) until the concentrations are equalized, as in part (b) of Figure 11.3. This tendency is called osmosis<\/strong>. In osmosis, the solute remains in its original side of the system; only solvent molecules move through the semipermeable membrane. In the end, the two sides of the system will have different volumes. Because a column of liquid exerts a pressure, there is a pressure difference \u03a0 on the two sides of the system that is proportional to the height of the taller column. This pressure difference is called the osmotic pressure<\/strong>, which is a colligative property.\r\n\r\n[caption id=\"attachment_7580\" align=\"aligncenter\" width=\"600\"]\"\" Figure 11.3 Osmosis. (a) Two solutions of differing concentrations are placed on either side of a semipermeable membrane. (b) When osmosis occurs, solvent molecules selectively pass through the membrane from the dilute solution to the concentrated solution, diluting it until the two concentrations are the same. The pressure exerted by the different height of the solution on the right is called the osmotic pressure.[\/caption]\r\n\r\nThe osmotic pressure of a solution is easy to calculate:\r\n

    [latex]\\Pi = MRT[\/latex]<\/p>\r\nwhere \u03a0 is the osmotic pressure of a solution, M<\/em> is the molarity of the solution, R<\/em> is the ideal gas law constant, and T<\/em> is the absolute temperature. This equation is reminiscent of the ideal gas law we considered in Chapter 6 \"Gases\"<\/a>.\r\n

    \r\n

    Example 11.5<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat is the osmotic pressure of a 0.333 M solution of C6<\/sub>H12<\/sub>O6<\/sub> at 25\u00b0C?\r\n\r\nSolution<\/em>\r\n\r\nFirst we need to convert our temperature to kelvins:\r\n

    [latex]T=25+273=298\\text{ K}[\/latex]<\/p>\r\nNow we can substitute into the equation for osmotic pressure, recalling the value for R<\/em>:\r\n

    [latex]\\Pi=(0.333\\text{ M})\\left(0.08205 \\dfrac{\\text{L}\\cdot\\text{atm}}{\\text{mol}\\cdot \\text{K}}\\right)(298\\text{ K})[\/latex]<\/p>\r\nThe units may not make sense until we realize that molarity is defined as moles per litre:\r\n

    [latex]\\Pi=(0.333\\hspace{2pt}\\dfrac{\\cancel{\\text{mol}}}{\\cancel{\\text{L}}})\\left(0.08205\\hspace{2pt} \\dfrac{\\cancel{\\text{L}}\\cdot\\text{atm}}{\\cancel{\\text{mol}}\\cdot \\cancel{\\text{K}}}\\right)(298\\text{ \\cancel{K}})[\/latex]<\/p>\r\nNow we see that the moles, litres, and kelvins cancel, leaving atmospheres, which is a unit of pressure. Solving, we get:\r\n

    \u03a0 = 8.14 atm<\/span><\/span><\/p>\r\nThis is a substantial pressure! It is the equivalent of a column of water 84 m tall.\r\n\r\nTest Yourself<\/em>\r\n\r\nWhat is the osmotic pressure of a 0.0522 M solution of C12<\/sub>H22<\/sub>O11<\/sub> at 55\u00b0C?\r\n\r\nAnswer<\/em>\r\n1.40 atm\r\n\r\n<\/div>\r\n<\/div>\r\nOsmotic pressure is important in biological systems because cell walls are semipermeable membranes. In particular, when a person is receiving intravenous (IV) fluids, the osmotic pressure of the fluid needs to be approximately the same as blood serum; otherwise bad things can happen. Figure 11.4 \"Osmotic Pressure and Red Blood Cells\" shows three red blood cells: (a) a healthy red blood cell; (b) a red blood cell that has been exposed to a lower concentration than normal blood serum (a so-called hypotonic<\/em> solution) so that the cell has plumped up as solvent moves into the cell to dilute the solutes inside; and (c) a red blood cell exposed to a higher concentration than normal blood serum (hypertonic<\/em>) such that water leaves the red blood cell and it collapses onto itself. Only when the solutions inside and outside the cell are the same (isotonic<\/em>) will the red blood cell be able to do its job.\r\n\r\n[caption id=\"attachment_7583\" align=\"aligncenter\" width=\"600\"]\"\" Figure 11.4 \"Osmotic Pressure and Red Blood Cells.\" (a) This is what a normal red blood cell looks like. (b) When a red blood cell is exposed to a hypotonic solution, solvent goes through the cell membrane and dilutes the inside of the cell. (c) When a red blood cell is exposed to a hypertonic solution, solvent goes from the cell to the surrounding solution, diluting the hypertonic solution and collapsing the cell. Neither of these last two cases is desirable, so IV solutions must be isotonic with blood serum to not cause deleterious effects.[\/caption]\r\n\r\nOsmotic pressure is also the reason you should not drink seawater if you\u2019re stranded in a lifeboat on an ocean; seawater has a higher osmotic pressure than most of the fluids in your body. You can<\/em> drink the water, but ingesting it will pull water out of your cells as osmosis works to dilute the seawater. Ironically, your cells will die of thirst, and you will also die. (It is OK to drink the water if you are stranded on a body of freshwater, at least from an osmotic pressure perspective.) Osmotic pressure is also thought to be important \u2014 in addition to capillary action \u2014 in getting water to the tops of tall trees.\r\n

    \r\n

    Key Takeaways<\/p>\r\n\r\n<\/header>\r\n

    \r\n