16.10 道德

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  毒性

  

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  How do you measure physical parameters of solutions?
  ::您如何测量解决方案的物理参数?

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  For many purposes, the use of is very convenient. However, when we want to know the of solute present in situations where there are temperature changes, molarity won’t work. The volume of the solution will change somewhat with temperature, enough to make accurate data observations and calculations in error . Another parameter is needed, one not affected by the temperature of the material we are studying.
  ::在许多方面,使用溶液非常方便。 但是,当我们想知道温度变化情况下溶液的存在时,摩尔度是行不通的。 溶液的量随温度而变化,足以进行准确的数据观测和错误计算。 需要另一个参数,一个不受我们研究的材料温度影响。

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  Molality
  ::毒性

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  A final way to express the concentration of a solution is by its molality. The molality ( m ) of a solution is the moles of solute divided by the kilograms of solvent . A solution that contains 1.0 mol of NaCl dissolved into 1.0 kg of water is a “one-molal” solution of sodium chloride. The symbol for molality is a lower-case m written in italics.
  ::表示溶液集中的最后一种方式是其软性。溶液的软性(m)是溶液的摩尔(m)除以千克溶剂。含有1.0毫摩尔的纳氏溶解成1.0千克水的溶液是氯化钠的“单摩尔”溶液。软性符号用斜体写成的小体m。

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  $$\begin{array}{r}\text{Molality}(m)=\frac{\text{moles of solute}}{\text{kilograms of solvent}}=\frac{\text{mol}}{\text{kg}}\end{array}$$

  
  ::闪度(m) =溶剂溶剂溶剂溶性树脂摩尔千克的摩尔值(m) =

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  Molality differs from molarity only in the denominator. While molarity is based on the liters of solution, molality is based on the kilograms of solvent. Concentrations expressed in molality are used when studying related to and temperature changes. Molality is used because its value does not change with changes in temperature. The volume of a solution, on the other hand, is slightly dependent upon temperature.
  ::软性与微量分母不同,虽然软性以溶液的公升数为基础,但软性则以溶剂的公斤为基础,在研究与温度变化和温度变化有关时,使用以软性表示的浓度。使用软性是因为其值不会随温度变化而变化。另一方面,溶液的体积则略微取决于温度。

   

   

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  Sample Problem: Calculating Molality
  ::样本问题:计算时间

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  Determine the molality of a solution prepared by dissolving 28.60 g of glucose (C ₆ H ₁₂ O ₆ ) into 250. g of water.
  ::确定将28.60克葡萄糖(C6H12O6)溶解成250克水的溶液的软性。

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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 solute= 28.60 g C ₆ H ₁₂ O ₆
    ::质量溶液= 28.60克 C6H12O6
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  • mass solvent = 250. g = 0.250 kg
    ::质量溶剂 = 250.g = 0.250公斤
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  • C ₆ H ₁₂ O ₆ = 180.18 g/mol
    ::C6H12O6 = 180.18克/摩尔

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  Unknown
  ::未知

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  • molality = ? m
    ::软性= 毫

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  Convert grams of glucose to moles and divide by the mass of the water in kilograms.
  ::将葡萄糖克转换成摩尔,以水质量除以公斤。

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  Step 2: Solve.
  ::步骤2:解决。

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  $$\begin{array}{rl}28.60{\text{ g C}}_6{\text{H}}_{12}{\text{O}}_6\times \frac{1{\text{ mol C}}_6{\text{H}}_{12}{\text{O}}_6}{180.18{\text{ g C}}_6{\text{H}}_{12}{\text{O}}_6}& =0.1587{\text{ mol C}}_6{\text{H}}_{12}{\text{O}}_6\\ \frac{0.1587{\text{ mol C}}_6{\text{H}}_{12}{\text{O}}_6}{0.250{\text{ kg H}}_2\text{O}}& =0.635m{\text{C}}_6{\text{H}}_{12}{\text{O}}_6\end{array}$$

  
  ::28.60克 C6H12O6x1摩尔 C6H12O6180.18克 C6H12O6=0.1587摩尔 C6H12O60.1587摩尔 C6H12O60.1587摩尔 C6H12O60.250千克 H2O=0.635米 C6H12O6

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

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  The answer represents the moles of glucose per kilogram of water and has three .
  ::答案代表每公斤水的葡萄糖摩尔数,

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  Molality and molarity are similar in value for dilute aqueous solutions because the density of those solutions is relatively close to 1.0 g/mL. This means that 1.0 L of solution has nearly a mass of 1.0 kg. As the solution becomes more concentrated , its density will not be as close to 1.0 g/ml and the molality value will be different than the molarity. For solutions with solvents other than water, the molality will be very different than the molarity. Make sure that you are paying attention to which quantity is being used in a given problem.
  ::溶液的密度相对接近1.0克/毫升,因此溶液的密度与稀释水溶液的值相似。这意味着1.0升溶液的重量接近1.0千克。当溶液更加集中时,其密度不会接近1.0克/毫升,而摩尔性值将不同于摩尔性。对于除水以外的溶剂的溶液,摩尔性与摩尔性将大不相同。请确保您注意某个问题所使用的数量。

   

   

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  Summary
  ::摘要

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  • The calculation of molality is described.
    ::报告叙述了对软性的计算。

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  Review
  ::回顾

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  1. What is the difference between molarity and molality? How are they similar?
     ::摩尔和摩尔之间有什么区别?它们是如何相似的呢?
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  2. Is molality affected by changes in temperature?
     ::软性是否受到温度变化的影响?
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  3. Why would the molality and molarity of a dilute aqueous solution be similar, but the molality and molarity of a highly concentrated aqueous solution not be similar?
     ::为什么稀释的水溶液的软质和纯度相似,但高度集中的水溶液的软质和纯度却不同呢?