16.14 冻结点萧条

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  冻结点萧条

  

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  Why salt icy roads?
  ::为什么是盐湿路?

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  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 (NaCl) and either calcium chloride (CaCl ₂ ) or magnesium chloride (MgCl ₂ ) are used most frequently, either alone or in a . Sodium chloride is the least expensive option, but is less effective because it only dissociates into two instead of three.
  ::协调性特性具有实际用途,例如在冷天气候下对道路加盐。通过对冰冷道路加盐,冰融点减少,冰融化速度加快,使驾驶更加安全。氯化钠(NaCl)和氯化钙(CaCl2)或氯化镁(MgCl2)的使用最为频繁,要么单独使用,要么在一个冷天气候中使用。氯化钠(MgCl2)是最便宜的选择,但效率较低,因为它只能分解为二,而不是三。

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  Freezing Point Depression
  ::冻结点萧条

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  Figure shows the phase diagram for a pure solvent and how it changes when a solute is added to it. The solute lowers the of the solvent resulting in a lowering of the 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 $\begin{array}{r}\mathrm{\Delta }{T}_f\end{array}$ .
  ::图中显示了纯溶剂的阶段图,以及溶液添加到溶液时的变化方式。溶液降低溶剂的溶剂,导致溶液点比溶剂低。冷点抑郁是纯溶剂冷点与溶液冷点之间的温度差异。在图中,冷点抑郁由 Tf 表示。

  

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  The vapor pressure of a solution (blue) is lower than the vapor pressure of a pure solvent (pink). As a result, the freezing point of a solvent decreases when any solute is dissolved into it.
  ::溶液(蓝色)的蒸气压力低于纯溶剂(粉末)的蒸气压力,因此溶剂的冻结点在溶解任何溶液时会减少。

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  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 make the hexagonally-shaped network of molecules that characterizes the . By dissolving a solute into the 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. 
  ::当纯溶剂冻结时,其颗粒会随着分子之间运行的分子间力的永久化而变得更为有序。在水中,使六角形分子网络成为溶剂的特性。通过溶解溶液,这种定序过程就会中断。因此,必须从溶液中去除更多的能量,才能冻结它,而溶液的冷点比纯溶剂低。

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  The magnitude of the freezing point depression is directly proportional to the of the solution. The equation is:
  ::冷点抑郁的程度与溶液直接成正比。 等式是 :

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  $$\begin{array}{r}\mathrm{\Delta }{T}_f=K_f\times m\end{array}$$

  
  ::Tf=Kf×m

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  The proportionality constant, $\begin{array}{r}{K}_f\end{array}$ , 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  $\begin{array}{r}{K}_f\end{array}$ is -1.86°C/ m . So the freezing temperature of a 1-molal aqueous solution of any nonvolatile molecular solute is -1.86°C. Every solvent has a unique molal freezing-point depression constant. These are shown in Table , along with a related value for the boiling point called _{$\begin{array}{r}{K}_b\end{array}$ .}
  ::相称性常数 Kf 被称为 摩尔冷点压低常数。 它等于非挥发分子溶液1- 摩洛溶液的冷点变化。 对于水, Kf 值为 - 1.86°C/m。 因此, 任何非挥发分子溶液1- 摩洛溶液的冷温为 - 1.86°C。 每个溶剂都有一个独特的摩洛冷点压低常数。 这些常数在表格中显示,以及沸点的相关值Kb。

  Molal Freezing-Point and Boiling-Point Constants

  Solvent	Normal freezing point (°C)	Molal freezing-point depression constant, K f (°C/ m )	Normal boiling point (°C)	Molal boiling-point elevation constant , Kb (°C/ m )
  Acetic	16.6	-3.90	117.9	3.07
  Camphor	178.8	-39.7	207.4	5.61
  Naphthalene	80.2	-6.94	217.7	5.80
  Phenol	40.9	-7.40	181.8	3.60
  Water	0.00	-1.86	100.00	0.512

   

   

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  Sample Problem: Freezing Point of a Nonelectrolyte
  ::抽样问题:非电气化冷冻点

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  Ethylene glycol (C ₂ H ₆ O ₂ ) is a molecular that is used in many commercial anti-freezes. 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. g of ethylene glycol in 500. g of water.
  ::乙二醇(C2H6O2)是一种分子,用于许多商用反冻物,在汽车散热器中使用乙烯甘醇的水溶液,以降低其冷点,从而防止散热器中的水被冷冻。计算400克乙二醇溶液在500克水中的冷点。

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  Step 1: List the known quantities and plan the problem.
  ::第1步:列出已知数量并规划问题。

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  Known
  ::已知已知

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  • mass C ₂ H ₆ O ₂ = 400. g
    ::质量 C2H6O2 = 400.g
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  • C ₂ H ₆ O ₂ = 62.08 g/mol
    ::C2H6O2 = 62.08克/摩尔
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  • mass H ₂ O = 500.0 g = 0.500 kg
    ::H2O = 500.0 克 = 0.500 公斤
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  • $\begin{array}{r}{\text{mass H}}_2\text{O}=500.0\text{g}=0.500\text{kg}\end{array}$
    ::H2O=500.0克=0.500千克
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  • $\begin{array}{r}{K}_f({\text{H}}_2\text{O})=-{1.86}^{\circ }\text{C}/m\end{array}$
    ::Kf(H2O)1.86C/m

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  Unknown
  ::未知

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  • $\begin{array}{r}{T}_f\text{of solution}=?{}^{\circ }\text{C}\end{array}$
    ::解答办法的Tf=? * * * * * * * * * * C * * * = * = * = * = * = * = * = * * C * * * = * = * = * = * = * = * = * = * = = * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * ** * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

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  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.
  ::这是一个三步问题。 首先,计算乙烯甘醇的摩尔。 然后计算溶液的体积。 最后,计算冷点抑郁。

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  Step 2: Solve.
  ::步骤2:解决。

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  $$\begin{array}{rl}400.{\text{ g C}}_2{\text{H}}_6{\text{O}}_2\times \frac{1{\text{ mol C}}_2{\text{H}}_6{\text{O}}_2}{62.08{\text{ g C}}_2{\text{H}}_6{\text{O}}_2}& =6.44{\text{ mol C}}_2{\text{H}}_6{\text{O}}_2\\ \frac{6.44{\text{ mol C}}_2{\text{H}}_6{\text{O}}_2}{0.500{\text{ kg H}}_2\text{O}}& =12.9m{\text{C}}_2{\text{H}}_6{\text{O}}_2\\ \mathrm{\Delta }{T}_f=K_f\times m=-{1.86}^{\circ }\text{C/}m\times 12.9m& =-{24.0}^{\circ }\text{C}\\ T_f& =-{24.0}^{\circ }\text{C}\end{array}$$

  
  ::400克 C2H6O2x1 mol C2H6O262.08克 C2H6O2=6.44 摩尔 C2H6O26.44摩尔 C2H6O20.500千克 H2O=12.9毫升 C2H6O2Tf=Kfxm1.86C/mx12.9毫升

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  The normal freezing point of water is 0.0°C. Therefore, since the freezing point decreases by 24.0°C, the freezing point of the solution is -24.0°C.
  ::因此,由于冷点减少24.0°C,溶液的冷点为-24.0°C。

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  Step 3: Think about your result.
  ::步骤3:想想你的结果。

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  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 in the result.
  ::水的冷点大量减少,防止散热器在冷冻时因水膨胀而受损,结果有三个。

    

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   Keeping roads clear of snow and ice in winter conditions isn't easy! The roads obviously can't be heated, so is ice and snow melted away? See if you can figure it out in this simulation: 
  ::冬季的冰雪和冰雪远离道路并不容易! 道路显然无法加热, 冰雪也会融化吗?

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  Summary
  ::摘要

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  • Freezing point depression is defined.
    ::冻结点抑郁症的定义已经界定。
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  • Calculations involving freezing point depression are described.
    ::介绍了涉及冷点抑郁症的计算。

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  Review
  ::回顾

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  1. How does a solute affect the freezing of water?
     ::溶液如何影响水的冻结?
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  2. How many moles of glucose would be needed to lower the freezing point of one kg of water 3.72°C?
     ::要降低一千克3.72°C水的冰点,需要多少颗葡萄糖的摩尔?
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  3. How many moles of NaCl would be needed to produce the same amount of lowering of temperature?
     ::要产生同样的温度降低量,需要多少纳克尔的摩尔?