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2011-09-20 16:43:00   来源：作文地带    双击单词自动翻译

　　Colligative Properties　　Properties of solutions that depend on the number of solute particles present per solvent molecule are called colligative properties. The concentration of solute in a solution can affect various physical properties of the solvent including its freezing point, boiling point, and vapor pressure. For the SAT II you will only need to be familiar with the first two.　　Freezing Point Depression　　The freezing point of a substance is defined as the temperature at which the vapor pressure of the solid and the liquid states of that substance are equal. If the vapor pressure of the liquid is lowered, the freezing point decreases.Why is a solution’s freezing point depressed below that of a pure solvent? The answer lies in the fact that molecules cluster in order to freeze. They must be attracted to one another and have a spot in which to cluster; if they act as a solvent, solute molecules get in the way and prevent them from clustering tightly together. The more ions in solution, the 美国GREater the effect on the freezing point. We can calculate the effect of these solute particles by using the following formula:

　　DTf = Kf msolute i

　　whereDTf = the change in freezing pointKf = molal freezing point depression constant for the substance (for water = 1.86oC/m)m = molality of the solutioni = number of ions in solution (this is equal to 1 for covalent compounds and is equal to the number of ions in solution for ionic compounds)　　Boiling Point Elevation　　As you learned earlier in this chapter, the boiling point of a substance is the temperature at which the vapor pressure equals atmospheric pressure. Because vapor pressure is lowered by the addition of a nonvolatile solute, the boiling point is increased. Why? Since the solute particles get in the way of the solvent particles trying to escape the substance as they move around faster, it will take more energy for the vapor pressure to reach atmospheric pressure, and thus the boiling point increases. We can calculate the change in boiling point in a way that’s similar to how we calculate the change in freezing point:

　　DTb = Kbmsolutei

　　whereKb = molal boiling point elevation constant (for water = 0.51 C/m) Now try a problem that deals with freezing point depression and boiling point elevation.　　Example　　Calculate the freezing point and boiling point of a solution of 100 g of ethylene glycol (C2H6O2) in 900 g of water.　　Explanation　　Calculate molality:

　　Freezing point depression = (m)(Kf)(i)

　　Tf = (1.79)(1.86)(1) = 3.33oC

　　Freezing point = 0oC - 3.33oC = -3.33oC

　　Boiling point elevation = (m)(Km)(i)

　　Tb = (1.79)(0.51)(1) = 0.91oC

　　Boiling point = 100oC + 0.91oC = 100.91oC

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