16.9 多元性

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  偶数

  

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  How many molecules can be found in a reaction?
  ::在反应中可以找到多少分子?

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  Chemists deal with amounts of molecules every day. Our reactions are described as so many molecules of A reacting with so many molecules of compound B to form so many molecules of compound C. When we determine how much reagent to use, we need to know the number of molecules in a given volume of the reagent. only tell us the number of grams, not molecules. A 100 mL solution of 2% NaCl will have a very different number of molecules than a 2% solution of CsCl. So we need another way to talk about numbers of molecules.
  ::化学家每天处理大量的分子。 我们的反应被描述为 如此多的A分子 与如此多的化合物B分子反应 形成如此多的化合物C分子。 当我们确定需要使用多少试剂时, 我们需要知道试剂量的分子数量。 只需告诉我们克数, 而不是分子数。 2% NaCl 的100毫升溶液将拥有非常不同的分子数量, 而不是 CsCl 的2%溶液。 所以我们需要另一种方式来谈论分子数量 。

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  Molarity
  ::偶数

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  Chemists primarily need the of solutions to be expressed in a way that accounts for the number of particles that react according to a particular chemical equation . Since percentage measurements are based on either mass or volume, they are generally not useful for . A concentration unit based on moles is preferable. The molarity (M) of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, you divide the moles of solute by the volume of the solution expressed in liters.
  ::化学分子主要需要以计算根据特定化学方程式反应的粒子数量的方式表达解决方案。由于百分比测量基于质量或体积,因此一般不起作用。基于摩尔的浓度单位比较可取。溶液的摩尔度(M)是溶溶于一升溶液的摩尔数。为了计算溶液的摩尔性,您将溶液的摩尔除以用立方体表示的溶液量。

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  $$\begin{array}{r}\text{Molarity (M)}=\frac{\text{moles of solute}}{\text{liters of solution}}=\frac{\text{mol}}{\text{L}}\end{array}$$

  
  ::多元性(M) = 溶解溶解溶解溶解物溶性体=molL的摩擦

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  Note that the volume is in liters of solution and not liters of solvent . When a molarity is reported, the unit is the symbol M and is read as “molar”. For example a solution labeled as 1.5 M NH ₃ is read as “1.5 molar ammonia solution”.
  ::请注意,体积是溶液的升数,而不是溶剂的升数,在报告一种溶液时,单位是M的符号,被读作“molar”,例如,标为1.5 M NH3的溶液被读为“1.5 摩尔氨溶液”。

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  Sample Problem: Calculating Molarity
  ::样本问题:计算平行性

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  A solution is prepared by dissolving 42.23 g of NH ₄ Cl into enough water to make 500.0 mL of solution. Calculate its molarity.
  ::将42.23克NH4Cll溶解到足够的水中,使500.0毫升溶解,从而形成一种解决办法。计算其纯度。

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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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  $$\begin{array}{rlrl}& \underset{\_}{\text{Known}}& & \underset{\_}{\text{Unknown}}\\ & \text{mass}=42.23\text{g}N{H}_{4}Cl& & \text{molarity}=?\text{ M}\\ & \text{molar mass}N{H}_{4}Cl=53.50\text{g}/\text{mol}\\ & \text{volume solution}=500.0\text{mL}=0.5000\text{L}\end{array}$$

  
  ::已知量=42.23 g NH4Clmolity =?mmorlar质量NH4Cl=53.50 g/mol 容量溶液=500.0 mL=0.5000 L

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  The mass of the ammonium chloride is first converted to moles. Then the molarity is calculated by dividing by liters. Note the given volume has been converted to liters.
  ::氯化铵的质量首先转换为摩尔。然后用升数来计算摩尔性。请注意,给定的体积已经转换为升。

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  Step 2: Solve.
  ::步骤2:解决。

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  $$\begin{array}{rl}42.23\text{ g }N{H}_{4}Cl\times \frac{1\text{ mol }N{H}_{4}Cl}{53.50\text{ g }N{H}_{4}Cl}& =0.7893\text{ mol }N{H}_{4}Cl\\ \frac{0.7893\text{ mol }N{H}_{4}Cl}{0.5000\text{ L}}& =1.579\text{ M}\end{array}$$

  
  ::42.23 g NH4Clx1 mol NH4Cl53.50 g NH4Cl=0.7893 mol NH4Cl0.7893 mol NH4Cl0.7893 mol NH4Cl0.7893 mol NH4Cl0.50000 L=1.579 M

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

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  The molarity is 1.579 M, meaning that a liter of the solution would contain 1.579 mol NH ₄ Cl. Four are appropriate.
  ::摩尔度为1.579MM,这意味着溶液的一升将包含1.579 mol NH4Cl。

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  In a laboratory situation, a chemist must frequently prepare a given volume of solutions of a known molarity. The task is to calculate the mass of the solute that is necessary. The molarity equation can be rearranged to solve for moles, which can then be converted to grams. See sample problem 16.3.
  ::在实验室情况下,化学家必须经常准备已知的摩尔度的一定数量的解决方案。 任务是计算必要的溶液质量。 摩尔度方程式可以重新排列, 用于解决摩尔, 然后可以转换成克。 见样本问题 16. 3。

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  Sample Problem:
  ::问题:

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  A chemist needs to prepare 3.00 L of a 0.250 M solution of potassium permanganate (KMnO ₄ ). What mass of KMnO ₄ does she need to make the solution?
  ::化学家需要准备高锰酸钾0.250M溶液的3.00升高锰酸钾溶液(KMNO4)。

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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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  • molarity = 0.250 M
    ::月度=0.250米
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  • volume = 3.00 L
    ::体积=3.00升
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  • KMnO ₄ = 158.04 g/mol
    ::KMnO4 = 158.0.04克/摩尔

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  Unknown
  ::未知

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  • mass KMnO ₄ = ? g
    ::质量 KMnO4 =? g

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  Moles of solute is calculated by multiplying molarity by liters. Then, moles is converted to grams.
  ::溶液的摩尔用升数乘以摩尔来计算。然后,将摩尔转换成克。

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  Step 2: Solve.
  ::步骤2:解决。

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  $$\begin{array}{rl}{\text{mol KMnO}}_4=0.250{\text{ M KMnO}}_4\times 3.00\text{ L}& =0.750{\text{ mol KMnO}}_4\\ 0.750{\text{ mol KMnO}}_4\times \frac{158.04{\text{ g KMnO}}_4}{1{\text{ mol KMnO}}_4}& =119{\text{ g KMnO}}_4\end{array}$$

  
  ::mol KMnO4=0.250 M KMnO4×3.00 L=0.750 mol KMnO4×3.00 L=0.750 mol KMnO40.750 mol KMnO4×158.04 g KMnO41 mol KMnO4=119 g KMnO4

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

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  When 119 g of potassium permanganate is dissolved into water to make 3.00 L of solution, the molarity is 0.250 M.
  ::当119克高锰酸钾溶解到水中以达到3.00升溶液时,月度为0.250毫升。

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  Watch a video of molarity calculations:
  ::观看微量计算视频:

   

   

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  Summary
  ::摘要

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  • Calculations using the concept of molarity are described.
    ::本报告介绍了使用货币概念的计算方法。

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  Review
  ::回顾

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  1. What does M stand for?
     ::M代表什么?
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  2. What does molarity tell us that percent solution information does not tell us?
     ::摩尔蒂告诉我们什么 百分率的解决方案信息没有告诉我们?
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  3. What do we need to know about a molecule in order to carry out molarity calculations?
     ::为了进行摩尔度计算,我们需要知道什么有关分子的信息?