经过分析，习题“实验室欲配制250mL0.2moloL-1的碳酸钠溶液，回答下列问题：可供选择的仪器：①量筒②胶头滴管③托盘天平④药匙⑤烧瓶⑥烧杯（1）通过计算可知，应用托盘天平称取____g碳酸钠晶体（Na2CO3o10H2...”主要考察你对“配制一定物质的量浓度的溶液”
等考点的理解。
因为篇幅有限，只列出部分考点，详细请访问。
配制一定物质的量浓度的溶液
与“实验室欲配制250mL0.2moloL-1的碳酸钠溶液，回答下列问题：可供选择的仪器：①量筒②胶头滴管③托盘天平④药匙⑤烧瓶⑥烧杯（1）通过计算可知，应用托盘天平称取____g碳酸钠晶体（Na2CO3o10H2...”相似的题目：
实验室配制500mL&0.500moloL-1的NaCl溶液，有如下操作步骤：①把称量好的NaCl晶体放入小烧杯中，加适量蒸馏水溶解；②把①所得溶液小心转入500mL容量瓶中；③继续向容量瓶中加蒸馏水至液面距刻度1～2cm处，改用胶头滴管小心滴加蒸馏水至溶液凹面底部与刻度线相切；④用少量蒸馏水洗涤烧杯和玻璃棒2～3次，每次洗涤的液体都小心转入容量瓶，并轻轻摇匀；⑤将容量瓶塞塞紧，充分摇匀．请填写下列空白：（1）操作步骤的正确顺序为（填序号）&&&&．（2）本实验必须用到的玻璃仪器有&&&&，&&&&，&&&&，&&&&．（3）某同学观察液面的情况如右图所示，对所配溶液浓度将有何影响？&&&&（选填“偏高”、“偏低”或“无影响”）．（4）若没有进行操作步骤④，对所配溶液浓度将有何影响？&&&&．（选填“偏高”、“偏低”或“无影响”）．（5）若实验过程中出现如下情况，应如何处理？①加蒸馏水时不慎超过了刻度线：&&&&．②向容量瓶中转移溶液时（实验步骤②）不慎有液滴掉在容量瓶外面：&&&&．
欲配制100ml&1.0mol/L&&Na2SO4溶液，正确的方法是&&&&①将14.2g&Na2SO4溶于100ml水中②将32.2g&Na2SO4o10H2O溶于少量水中，再用水稀释至100ml③将20ml&5.0mol/L&Na2SO4溶液用水稀释至100ml．①②②③①③①②③
精确配制250mL一定物质的量浓度的NaOH&溶液，下克实验操作正确的是&&&&选择仪器时，使用500mL容量瓶将称量好的氢氧化钠固体放入容量瓶中，加入少量水溶解在烧杯中溶解氢氧化钠固体后，立即将所得溶液注入容量瓶中将烧杯中的氢氧化钠溶液注入未经干燥的洁净容量瓶中
“实验室欲配制250mL0.2moloL-...”的最新评论
该知识点好题
1下列实验操作中，错误的是（　　）
2下列说法正确的是（　　）
3实验室用98%的浓硫酸（密度为1.84g/mL）配制0.5mol/L硫酸溶液500mL，不需要使用的仪器是（　　）
该知识点易错题
1（2009o广东）超细氧化铝是一种重要的功能陶瓷原料．（1）实验室常以NH4Al（SO4）2和NH4HCO3为原料，在一定条件下先反应生成沉淀NH4AlO（OH）HCO3该沉淀高温分解即得超细Al2O3．NH4AlO（OH）HCO3热分解的化学反应方程式&&&&．（2）NH4Al（SO4）2o12H2O的相对分子质量为453．欲配制100mLPH为2、浓度约为0.1mol-1的NH4Al（SO4）2溶液，配制过程为①用托盘天平称量NH4Al（SO4）2o12H2O固体&&&&g；②将上述固体置于烧杯中&&&&．（3）在0.1moloL-1NH4Al（SO4）2溶液中，铝各形态的浓度（以Al3+计）的对数（lgc）随溶液PH变化的关系见下图①用NaOH溶液调节（2）中溶液PH至7，该过程中发生反应的离子方程式有&&&&．②请在答题卡的框图中，画出0.01moloL-1NH4Al（SO4）2溶液中铝各形态的浓度的对数lgc随溶液PH变化的关系图，并进行必要的标注．
2下列说法正确的是（　　）
3实验中需0.1mol/L的Na2CO3溶液950mL，用容量瓶配制时应称取Na2CO3oH2O粉末的质量为（　　）
欢迎来到乐乐题库，查看习题“实验室欲配制250mL0.2moloL-1的碳酸钠溶液，回答下列问题：可供选择的仪器：①量筒②胶头滴管③托盘天平④药匙⑤烧瓶⑥烧杯（1）通过计算可知，应用托盘天平称取____g碳酸钠晶体（Na2CO3o10H2O）．（2）配制过程需要的仪器____（填序号），还缺少的仪器有____．（3）配制时，正确的操作顺序是（用字母表示，每个字母只能用一次）____；A．用30mL水洗涤烧杯2～3次，洗涤液均注入容量瓶B．用托盘天平准确称取所需的Na2CO3o10H2O晶体的质量，放入烧杯中，再加入少量水（约30mL），用玻璃棒慢慢搅动，使其充分溶解C．将已冷却的溶液沿玻璃棒注入250mL的容量瓶中D．将容量瓶盖紧，振荡，摇匀E．改用胶头滴管加水，使溶液凹液面恰好与刻度相切F．继续往容量瓶内小心加水，直到液面接近刻度1～2cm处（4）若出现如下情况，对所配溶液浓度将有何影响（填“偏高”、“偏低”或“无影响”）．若没有进行A操作____；若定容时俯视刻度线____．”的答案、考点梳理，并查找与习题“实验室欲配制250mL0.2moloL-1的碳酸钠溶液，回答下列问题：可供选择的仪器：①量筒②胶头滴管③托盘天平④药匙⑤烧瓶⑥烧杯（1）通过计算可知，应用托盘天平称取____g碳酸钠晶体（Na2CO3o10H2O）．（2）配制过程需要的仪器____（填序号），还缺少的仪器有____．（3）配制时，正确的操作顺序是（用字母表示，每个字母只能用一次）____；A．用30mL水洗涤烧杯2～3次，洗涤液均注入容量瓶B．用托盘天平准确称取所需的Na2CO3o10H2O晶体的质量，放入烧杯中，再加入少量水（约30mL），用玻璃棒慢慢搅动，使其充分溶解C．将已冷却的溶液沿玻璃棒注入250mL的容量瓶中D．将容量瓶盖紧，振荡，摇匀E．改用胶头滴管加水，使溶液凹液面恰好与刻度相切F．继续往容量瓶内小心加水，直到液面接近刻度1～2cm处（4）若出现如下情况，对所配溶液浓度将有何影响（填“偏高”、“偏低”或“无影响”）．若没有进行A操作____；若定容时俯视刻度线____．”相似的习题。实验室用Na2CO3o10H2O晶体配制250mL&0.1mol/L&Na2CO3溶液，回答下列问题：（1）用托盘天平称取Na2CO3o10H2O的质量为______g．（2）操作中所需仪器除托盘天平、烧杯、玻璃棒外，还需要______（填仪器名称）；（3）配制时，正确的操作顺序是（用字母表示，每个字母只能用一次）______；A．用蒸馏水洗涤烧杯及玻璃棒2-3次，洗涤液均注入容量瓶，振荡B．用托盘天平称量所需的Na2CO3o10H2O晶体，放入烧杯中，再加入少量水，用玻璃棒慢慢搅动，使其完全溶解C．将已冷却的Na2CO3溶液沿玻璃棒注入容量瓶中D．将容量瓶盖紧，摇匀E．改用胶头滴管加水，使溶液凹液面的最低点恰好与刻度线相切F．继续往容量瓶内小心加水，直到液面接近刻度线1-2cm处（4）若出现如下情况，对所配溶液的浓度将有何影响（填“偏高”、“偏低”或“无影响”）：①没有进行A操作______；②称量时，Na2CO3o10H2O已失去部分结晶水______；③若定容时俯视刻度线______；④定容摇匀时，发现液面下降，又加水至刻度线______；⑤容量瓶用蒸馏水洗净后没有干燥______．
（1）需要Na2CO3o10H2O的物质的量是：0.250L×0.1mol/L=0.025mol，质量为：0.025×286=7.15g，用托盘天平称取质量是7.2g，故答案是：7.2；（2）配制一定浓度的溶液需要的仪器有：托盘天平、烧杯、玻璃棒、胶头滴管、250mL容量瓶等，故答案是：250mL容量瓶；&&胶头滴管；&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& （3）配制步骤是：计算、称量、容积、冷却、转移、洗涤、定容，所以正确的操作顺序是B、C、A、F、E、D，故答案是：B、C、A、F、E、D；&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& （4）①配制溶液时必须相同，否则配制的溶液中溶质的物质的量减小，浓度偏低，故答案是：偏低；②由于晶体失去结晶水，称取的溶质的质量变大，溶质的物质的量变大，浓度偏高，故答案是：偏高；③进行定容时，若是俯视刻度线，导致配制的溶液体积偏小，浓度偏高，故答案是：偏高；④定容摇匀时，发现液面下降，又加水至刻度线，导致配制的溶液体积偏大，浓度偏低，故答案是：偏低；⑤容量瓶用蒸馏水洗净后没有干燥，不会影响配制结果，故答案是：无影响．
为您推荐：
（1）计算出需要Na2CO3o10H2O晶体的物质的量，根据摩尔质量计算出质量，注意托盘天平准确到0.1g；（2）根据配制一定物质的量浓度的溶液使用的仪器完成，注意容量瓶必须标明容积；（3）根据配制一定物质的量浓度溶液的操作步骤进行排序；（4）①没有洗涤，溶质的物质的量减小；②失去结晶水，称取的溶质质量偏大，物质的量偏大；③定容时俯视，导致配制的溶液体积减小；④加入蒸馏水，配制的溶液体积偏大；⑤容量瓶内的蒸馏水对配制结果没有影响．
本题考点：
配制一定物质的量浓度的溶液．
考点点评：
本题考查了配制一定物质的量浓度的溶液的配制方法及误差分析，涉及题量稍大，难度中等，可以根据所学知识完成．
扫描下载二维码Solutions Stoichiometry
Solution Stoichiometry
Key words:
solution, solute, and solvent
solubility
concentraion
Skills to develop
Expressing concentrations
Converting concentrations between various units: g/L, percent, weight
percentate, mole percent, mol/L (M), ppm, etc.
Calculating amounts of solute in a given volume of solution
Amount = Concentration * Volume
Applying this formula to solve titration problems
Preparing a solution of prescribed concentration
Solving any problem involving solution stoichiometry
Solution Stoichiometry
The topic solution stoichiometry deals with quantities
in chemical reactions taking place in solutions. Once you have mastered
this topic, you will be able to prepare solutions of desirable concentrations,
carry out chemical reactions using correct amounts of solutions, predict
amounts produced, and calculate yields. In order build your skills, you are
going to calculate concentrations and amounts of solute present in the
are homogeneous mixtures, the major
component being the solvent, and minor components being solute.
In terms of stoichiometry, we mainly deal with
and liquid solutions, but
properties solid solutions are emphasized in material sciences.
Solution Concentration
One of the most important properties of a solution is the concentration,
which is the amount of solute dissolved in a given volume or weight of
solution. Since there are different ways to represent the quantities,
concentration can be expressed in different units. Weight of solute per unit
volume (of solution), amount in mole per unit volume, weight percentage,
and mole percentage are some of the ways.
For the application in chemical stoichiometry, the unit of moles per litter
(mol/L) is the most convenient, but the real world uses other expressions
for concentration. We need the skill to convert from one type of units to
another, because we are part of the real world. Furthermore, we should
also have the ability to prepare a solution of desirable concentration,
and in this task, we use quantities measured in conventional units
(g, kg, L, m3 etc).
In an era of environmental concerns and pollution sensitivity brought
about by sensitive chemical analysis, the units ppm (part per million
1/106), ppb (part per billion 1/109, and
ppt (part per trillion 1/1012) are often used.
It is interesting to note that 0.1 percent is the same as 1000 ppm,
but the latter sounds more serious.
In a laboratory, Qem dissolved 11.2 g of sugar in water, and then
he poured the solution into a 250-mL volume flask. He added enough
water to make the solution exactly 250.00 mL. What is the concentration
of the solution he has prepared?
The molecular weight for sugar C12H22O11
is 342.0 g/mol. (Your should have the skill to calculate molecular weight.)
For solving chemical problems, the units mol/L is the most useful.
Thus, the concentration is:
11.2 g ------- ------- = 0.131 mol/L (M)
342.0 g 0.250 L
The concentration is 0.131 M
Discussion
We have to measure the density in order to express the concentration in
percentage and other units.
If the 0.131-M sugar solution has a density of 1.10 g/mL, what is the
concentration in mole percentage?
The total weight of the solution is 1.10*250 = 275 g.
Amount of sugar = (0.131 mol/L)(0.250 L) = 0.0327 mol.
(275 - 11.2)g
Amount of water = -------------- = 14.7 mol.
Thus, the mole percentage = --------------- x 100 = 0.22 %
14.7 + 0.0327
At this level, the concentration can be expressed as 2200 ppm by mole.
However, the concentration is 4.1 % by weight or 41000 ppm.
What is the weight percentage?
A well known fact is that at standard temperature and pressure, one mole
of ideal gas occupy 22.4 L. What is the concentration in M
of this ideal gas?
Concentration = ------ = 0.0446 M.
For a gas at 273 K, what is the pressure if the concentration is 1.0 M?
Solution Stoichiometry
The amount of solute in certain volume of solution is equal to the
volume (V) multiplied by the concentration (C).
Amount = C * V
If the units are included as part of your formulation or calculation,
you can derive the correct unit to express the amount.
Delivering a desirable amount of substance is best done by using volume.
For example, when the concentration is 0.1 M, 1.0 mL solution contains
1.0 micromole.
How much NaOH (formula weight = 40) is contained in 25.0 mL of a
solution whose concentration is 0.1234 M (same as mol/L).
Amount = 0.025 L * 0.1234 mol/L = 3.085x10-3 mol
= 0.1234 g
If 1.000 mL of this solution (0.1234 M) is diluted to 250.0 mL in a
volumetric flask, what is the concentration of the final solution?
How much NaOH is there in 0.1 mL of the diluted solution.
In problems involving neutralization such as the reaction equation:
2 HCl + Ca(OH)2 = CaCl2 + 2 H2O
one mole of HCl reacts with half a mole of Ca(OH)2.
Half mole of Ca(OH)2 is the equivalence to one mole of HCl.
Similarly for the reaction involving oxidation reactions,
KMnO4 + 5 FeCl2 + 8 HCl = KCl + MnCl2 + 5 FeCl3
One mole of KMnO4 is equivalent to 5 moles of
This reaction equation shows the requirement of 8 moles
of HCl, but in reality, KMnO4 will oxidize Cl-
into chlorine, Cl2. We used HCl for simplicity in balancing
the equation.
If the concentrations of the two solutions are
C1 and C2, and the volumes at the
equivalence points are V1 and V2
for solute 1 and 2 respectively, their amounts can be calculated.
Amount1 = C1 V1
Amount2 = C2 V2
If Amount1 is equivalent to Amount2,
then we have
C1 V1 = C2 V2
If Amount1 is equivalent to n Amount2,
then we have
C1 V1 = n C2 V2
These equations are useful for titration calculations. Since their derivation
is so trivial, you are expected to derive them whenever required.
The two examples below show the application of the above discussion.
If 25.00 mL HCl acid with a concentration of 0.1234 M is neutralized
by 23.45 mL of NaOH, what is the concentration of the base?
[NaOH] = 0.02500 L * 0.1234 mol/L /0.02345 = 0.1316 M mol
This type of calculation is used for titrations, an important technique
in a laboratory.
If 25.00 mL HCl acid with a concentration of 0.1234 M is neutralized
by 23.45 mL of Ca(OH)2, what is the concentration of the base?
[Ca(OH)2] = 0.5 * 0.02500 L * 0.1234 mol/L /0.02345 = 0.06508 M mol
This concentration is half of that of NaOH.
Discussion
[OH] = 0.1316 M in both cases of Example 6 and 5.
Solutions are used in the food we eat, the air we breathe, and the
medicine we use. Solutions are also everywhere around us in the
environment. Thus, there is a need to keep track of concentrations
and amounts of key substances in these solutions so that we are at
least aware of how much we eat, drink, or take.
Confidence Building Problems
You dissolved 10.0 g of sugar in 250 mL of water (the volume of a
cup of coffee). Calculate the weight percentage of sugar in
the solution. Assume the density of solution to be 1.0 g/mL.
Hint . . .
100 * 10.0/(250 + 10) = ? %
Calculate and estimate concentration.
You dissolved 10.0 g of sugar to make 250 mL of solution.
Calculate the concentration in M. (Molar mass of sugar
C12H22O11 = 342).
Hint . . .
? mol/L or M.
Exercise -
What is the concentration if you dilute the solution to 1.0 L?
You dissolved 13.0 g of NaCl to make 2.00 L of solution.
Calculate the molarity. Atomic wt: Na, 23.0; Cl, 35.5.
Hint . . .
------------- = ? mol/L
(23.0+35.5) g
Exercise -
What is the concentration if the water evaporated, and the volume
is reduced to 1.00 L?
You diluted 10.00 mL of 6.0 M H2SO4 to prepare
250.0 mL solution. Calculate the concentration of this solution.
Molar mass: H2SO4, 98.0
Hint . . .
Never pour water into concentrated sulfuric acid.
How much calcium carbonate Ca(HCO3)2 is present
in 24.0 L of tap water if analysis indicates that the tap water
contains 42.0 ppm Ca(HCO3)2?
Assume tap water density to be 1.00 g/mL.
Hint . . .
Apply the equation: Amount = C * V.
Ca(HCO3)2 is present in hard-water.
A stock sulfuric acid solution contains 98.0 % of H2SO4,
and its density is 1.840 g/mL. How many mL of this acid is required to
prepare 5.0 L of 2.0 M solution? Molar mass: H2SO4,
Hint . . .
100 g stock
5.0*2 = 10 mol H2SO4 ------
-----------
98 g H2SO4
Practical information -
Stock sulfuric acid is 18.4 M. Calculate it.
A bottle of nitric acid has a density of 1.423 g/mL, and
contains 70.9 % HNO3 by weight. What is the molarity?
Molar mass: HNO3 = 63.01
Hint . . .
Assume you have 1.00 L solution.
1000 mL solution 1.423 g
70.9 g HNO3
---------------- --------- --------------- ------------ = ? mol/L
100 g solution
63.01 g HNO3
Many problem solutions require assumptions. Make the appropriate
assumption in problem solving.
If you want to make 250.0 mL 0.100 M calcium chloride solution,
how many moles of CaCl2 will you need?
Hint . . .
Other units -
You need 25 mmole or 0.025 mol.
If you want to make 250.0 mL of 0.100 M calcium chloride solution,
how many grams of CaCl2 are needed? Molar mass: CaCl2,
111.1 g/mol.
Hint . . .
0.250 L -------
? g (CaCl2)
Prepare a solution of definite concentration.
What is the molarity of Na+ in a 0.123 M solution of
Hint . . .
Explain the dissolution as ionization:
Na2SO4 = 2 Na+ + SO42-
How many mL of 0.200 M aluminum chloride (AlCl3) solution will
contain 6.00 millimole of Cl- ions?
Hint . . .
Exercise...
How many mL will contain 6 millimole Al3+ ions?
How many mL of stock HNO3 solution (16.0 M) is required to make
400.0 mL 2.0 M solution?
Hint . . .
Excellent...
Prepare solution of certain cencentration by dilution.
How many mL of 0.321 M HCl solution is required to neutralize
5.0 mL 1.284 M KOH solution?
Hint . . .
Performing calculations of titration results.
If 10.0 mL HNO3 completely neutralize 25.0 mL of 0.351 M KOH,
calculate the molarity of HNO3.
Hint . . .
Further consideration -
What is the answer if sulfuric acid is used?
Given a solution containing 0.242 g of barium chloride, BaCl2,
how many mL of 0.0581 M H2SO4 will completely
precipitate the barium ions, Ba2+? Molar mass: BaCl2,
Hint . . .
Performing volumetric analysis.
A 0.5404 g sample containing NaOH and NaCl is dissolved, and
titrated with standard 0.1010 M H2SO4.
If 14.32 mL of the acid is required, calculate the percentage of
NaOH in the sample?
Molar mass: NaOH, 40.0.
Hint . . .
Convert mol H2SO4 to mol NaOH to weight NaOH to % NaOH
& CChieh@UWaterloo.caTo get the best deal on Tutoring, call 1-855-666-7440
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Concentration of a Solution
A solution is a homogenous mixture of two or more pure substances whose composition may be altered within certain limits. Though the solution is homogenous in nature, yet it retains the properties of its constituents.
Generally solution is composed of two components, solute and solvent. Such type of solution is known as binary solutions.
Solvent is that component in solution whose physical state is the same as that of the resulting solution while other component is called as solute. If the physical state of both component is same, than the component in excess is known as solvent and other one is called as solute. Each component in a binary solution can be in any physical state such as liquid, solid and gaseous state.
Some common type of solutions is as follows.&
Types of Solution&
Solute Solvent&
&Gaseous Solution&
&Sub limitationof Solid in G Camphor Vapour in $N_2$ gas.Water Vapours in air(mist),Mixture of gas ,air.
&Liquid&Gas&Liquid Solution&Solid&Liquid&Salt in Water, Sucrose in Water,alcohol in Water,$CO_2$ in waterAerated drink.& &Liquid&Gas&Solid Solution&Solid&Solid&Alloys, Adsorption of Gas on Metal ,Solution of Hydrogen in& Palladium.&Liquid
Finding the Concentration of a Solution
The concentration of a solution is the amount of solute dissolved in a known amount of the solvent or solution. Solution can be described as dilute or concentrated solution as per their concentration. A dilute solution has a very small quantity of solute while concentrate solution has a large quantity of solute in solution. Various concentration terms are as follows.1. Mass percentageIt may be defined as the number of parts of mass of solute per hundred parts by mass of solution. If WB is the mass of solute B and Wa is the mass of solvent A, thenMass % = $\frac{W_B}{W_A + W_B}$
x 100For example, the percent composition by mass of a 100 g sugar solution which contains 15 g sugar willbe$\frac{15g\ Sugar}{100g\ solution}$
x 100 = 15% Sugar solution2. Volume PercentIt can be represented as % v/v or % volume and used to prepare such solutions in which both components are in liquids state. It is the number of parts of by volume of solute per hundred parts by volume of solution. Therefore,Volume % = $\frac{volume\ of\ solute}{total\ volume\ of\ solution}$ x 100%For example, 20% solution of ethanol by volume means that 100 cm3 of the solution contains 20 cm3 of ethanol and 80cm3 of water.3. Mass-volume percentage (W/V %)It may be defined as the mass of solute present in 100cm3 of solution. For example, If 100 cm3 of solution contains 5g of sodium hydroxide, than the mass-volume percentage will be 5% solution. 4. Parts per million (ppm)The very low concentration of solute in solution can be expressed in ppm. It is the numbers of parts by mass of solute per million parts by mass of the solution. Parts per million (ppm) = $\frac{Mass\ of\ solute X 10^6}{Mass\ of\ solution}$
5. Mole Fraction (X)Mole fraction may be defined as the ratio of number of moles of one component to the total number of moles of all the components (solute and solvent) present in solution. It is denoted by letter x and the sum of all mole fractions in a solution always equals one. If a solution contains the components A and B and WA g of A and WB g of B are present in it, than the moles of A (nA) = WA/(MA);Moles of B(nB) = WB/(MB)Where MA and MB are molar mass of A and B respectively Total number of moles of A and B = nA + nBHence Mole fraction of A (xA) = nA /
nB Mole fraction of B (xB) = nB / nA +
nB xA + xB = (
)= 1 Mole fraction is not depends upon temperature and can be extended to solutions having more than two components. 6. Molarity (M)Molarity is most common unit for concentration of solution. It is defined as the number of gram mole of solute present in one litre or one dm3 of the solution or millimol of solute present in one mL of solution. Mathematically, it can be expressed asM = $\frac{Mole\ of\ solute}{Volume\ of\ solution(L)}$Since mole of solute = Mass of solute in gram/ Molar mass of soluteTherefore, M = Mass of solute in gram/ Molar mass of solute x Volume of solution (L)If
nB moles of solute B are present in V liter of solution.Molarity = $\frac{n_B}{V(L)}$OrM = $\frac{W_B X 1000}{M_B X V(ml)}$Where MB is the molar mass of solute and WB is the gram of solute present in V(ml) of solution. One molar solution is defined as the solution having unity Molarity. Such solutions have one mole of solute per liter of solution. The unit of Molarity is mol/L or mol dm3-. 7. Molality(m)The number of gram mole of the solute present in 1000 g of the solvent is known as molality of solution. It represented by letter ‘m’.Molality (m) = $\frac{Mole\ of\ solute}{Mass\ of\ solvent(kg)}$Or m = $\frac{Mass\ of\ solute(g) X 1000}{Molar\ mass\ of\ solute X mass\ of\ solvent(g)}$The unit of molality is mol/kg and it does not effect by temperature.8. Normality(N)The number of gram-equivalent of the solute present in one liter of solution is known as normality of solution. I number of milli-equivalents of solute present in one mL of solution is called as normality of that solution. It represented by letter ‘N’.Normality (N) = $\frac{gram\ equivalent\ of\ the\ solute}{volume\ of\ solution(L)}$A solution having normality equal to unity is known as normal solution. Such solution contains one gram equivalent of solute per dm3 of solution.
Calculating the Concentration of a Solution
Concentration of solution can be calculated in terms of normality, Molarity, molality or mole fraction.
Solved Examples
Question&1: Calculate the mole fractions of the components of the solution composed by 92 g glycerol and 90 g water? (M(water) = 18; M(glycerol) = 92)
&Moles of water = 90 g x&1 mol / 18 g = 5 mol waterMoles of glycerol = 92 g x 1 mol / 92 g = 1 mol&glycerolTotal moles in solution = 5 + 1 = 6 molMole fraction of water xwater&=&5 mol / 6 mol = 0.833Mole fraction of glycerol xglycerol&=&1 mol / 6 mol = 0.167&
Question&2: What will be the Molarity of solution when water is added to 10 g CaCO3 to make 100 mL of solution?
&10 g CaCO3&/&(100 g CaCO3&/ mol CaCO3) = 0.10 mol CaCO3100 mL x 1 L / 1000 mL = 0.10 LMolarity = Mole of solute / Volume of&solution (L)= 0.10 mol / 0.10 LT Molarity of given solution = 1.0 M&
Question&3: Calculate the molality of a solution containing 20 g of sodium hydroxide (NaOH) in 250 g of water?
&Moles of&sodium hydroxide= 20 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.2 mol NaOH250 gm = 0.25 kg of water (250 g water x 1 kg / 1000 g = 0.25 kg)Hence molality of solution = Mole of solute/ Mass of solvent (kg)= 0.2 mol / 0.25&kgor Molality(m) = 0.8M / kg or 0.8 m&
Question&4: Calculate the number of grams of copper sulphate (CuSO4)
needed to prepare 250.0 mL of 1.00 M CuSO4?
&Molarity = $\frac{Mass\ of\ copper\ sulphate\ (gm) X 1000}{Molar\ mass\ of\ copper\ sulphate\ X\ V (ml)}$&&&&&&&&&&&&&&&&&& & & & & & & & & &
Molar mass of copper sulphate = 159.6 g/molV =250 mlMolarity =2.00M&Hence Mass of copper sulphate (gm) =& $\frac{Molarity X Molar\ mass\ of\ copper\ sulphate\ X\ V (ml)}{1000}$&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& =&$\frac{2.00\ X\ 159.6\ X\ 250}{1000}$&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& = 79.8 gm of Copper sulphate&
Question&5: Calculate the normality of solution containing 296.4 grams of Ca(OH)2 in 600 ml of water.
&Normality = $\frac{mass\ of\ solute (gm) X 1000}{Equivalent\ mass\ of\ solute X V (ml)}$&&&&&&&&&&&&&&&& Equivalent mass of calcium hydroxide = Molar mass/ acidityMolar mass of calcium hydroxide= 74.1 amuAcidity =2 (as Ca(OH)2& contains 2 replaceable hydroxides) Therefore&Equivalent mass of calcium hydroxide = 74.1/2=37.05&Normality of solution =&$\frac{296.4 X
X 600}$= 13.3 N &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Question&6: Calculate the ppm of mercury in water in given sample contain 30 mg of Hg in 500 ml of solution.
&Parts per million = $\frac{Mass\ of\ solute X 10^6}{Mass\ of\ solution}$&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Mass of Hg = 30 mgMass of water = 500/1 = 500g = 50 x 104 mg(As Mass = Volume / density of water 1 g/ml)&Therefore ppm of mercury = $\frac{30mg\ of\ lead X 10^6}{50 X 10^4 of\ solution}$= 60 ppm of mercury&
Formula for Concentration of a Solution
Concentration of Ions in a Solution
In general, we are using ionic compounds in various reactions. Ionic compounds are good conductor in their aqueous solution due to the presence of ions in solution. Therefore in ionic compounds the determination of concentration of ions is more useful than concentration of solution. For determination of concentration of ions in solution, first we have to write balance chemical equation that show how these substances break down into ions. For example, the dissociation of potassium carbonate in aqueous can be written as follows.K2CO3 (s) $\to$ 2 K+(aq)+ CO32-(aq)Every one mole of K2CO3
produces two moles of K+ ions and two moles of carbonate ions in solution with no change in volume. Now if we have 0.40 M solution of potassium carbonate ( K2CO3
), the concentration of both ions. K+ and
CO32- can be calculated by using the coefficients in our balanced equation.Concentration of potassium ion, [K+] = 2 × [ K2CO3
] = 2 × 0.40 M = 0.80 MConcentration of carbonate ion [ CO32- ] = 1 × [
] = 1 × 0.40 M = 0.40 MLet’s calculate the concentration of ions in 0.30 M solution of Iron (III) nitrate. The balanced chemical equation of dissociation of Iron (III) nitrate in its aqueous sFe(NO3)3 $\to$ Fe3+(aq) + 3 NO3-(aq)
The coefficients from the balanced chemical equation may used to determine the concentration of the ions.Concentration of [ Fe3+ ] = 1× [
] = 1 × 0.30= 0.30 MConcentration of [ NO3- ] = 3× [
] = 3 × 0.30 = 0.90 MExample 1Calculate ion concentrations in a 3.00 L solution containing 17.1 g aluminium sulphate, Al2(SO4)3. (Atomic mass of Al =27.0 g/mol, O=16 g/mol, S =32.0 g/mol)Calculation :Molar mass of Aluminium sulphate = 2 × 27.0 +3× 32.0.0+12 × 16.0 = 342.0 g/molConcentration of
Al2(SO4)3 = Mass of
Al2(SO4)3 / Molar mass of
x V(L)= 17.1 /342.0 x 3 = 0.0166 MDissociation of aluminium sulphate in its aqueous solution can be written asAl2(SO4)3
$\rightarrow$ 2Al3+ (aq) + 3SO42- (aq)From balanced equaConcentration of [Al3+] = 2× [ Al2(SO4)3 ] = 2 × 0.0166 = 0.0333 M or 3.33 ×10-2MConcentration of [SO42-] = 3× [ Al2(SO4)3 ] = 3 × 0.0166
= 0.0498 M or 4.98×10-2MExample 2What is the concentration of sodium and sulphate ions in a 0.650 M solution of Na2SO4?Calculation:The dissociation of sodium sulphate in its aqueous solutiNa2SO4(s) $\rightarrow$ 2Na+(aq) + SO42-(aq) The concentration of sodium and sulphate ions will beConcentration of S [Na+] = 0.650 M Na2SO4 x 2Na+ /1
Na2SO4 = 1.3 M Na+Concentra [SO42-] = 0.650 M Na2SO4 x1SO42- /1
Na2SO4 = 0.650 M
Like Molarity is use to expressed the concentration of solution, formality is used to calculate the concentration of ions in solution. It is the number of formula weight units of solute present in one liter of solution. Formality is represented by letter ‘F’. The mass of one mole of a compound equals to the formula weight in grams. Formality is used for the determination of the number of moles of a compound from the number of moles of ions present in solutions of ionic compounds.F = $\frac{n Formula\ weight\ units}{ Volume\ of\ solution\ (L)}$
Determining the Concentration of a Solution Beer's Law
The properties of light can be described by wavelength and frequency.
On the basis of wavelength and frequency, light can be classified in various types. Out of these radiations, only visible light can be seen by human eyes. The wavelength of visible light lies between 400 nm to 750 nm. The visible spectrum is composed of seven colors, where each color associated with certain wavelength. When visible light falls on the surface of any substance, it absorbs and complimentary color transmitted. The color of object is due to complementary color which is just opposite to absorbed color. The mathematical relation between absorption of light with concentration of solution may be given by Beer-Lambert Law.
A = ebcWhere
,A = Absorbance of light at given wavelength<e
= Molar absorptivity(L mol-1 cm-1)b = Path length through the solution that the light has to travel (cm)c = Concentration of the solution (mol/liter)As the absorption of a particular color of light increases, the color of solution increases in that region of spectrum which is not absorbed by solution. The absorbance (A) is directly proportional to the concentration of solution, therefore increases with increasing the concentration of solution. For the determination of concentration of solution of an unknown sample, the absorption of a series of known solution is measured and plotted which is called as Beer’s law plot or calibration curve. This curve can be used for the determination of concentration of unknown sample by measuring the absorbance at same wavelength and path length.
Colorimetry is used for the determination of concentration of solution which based on the relation between absorption and path length. Since molar absorptivity is unique for each absorbing species and vary with wavelength, therefore any sum of path length and concentration will be same when multiplied will absorb the same amount of light. Absorbance is related to appearance of solution, therefore two solutions (A1 and A2) with same color will show same absorbance.A1 = A2From Beer-Lambert Law
e bc, e 1 =
e 2 (for same substance)Therefore b1c1 = b2c2Orb1c1 = b2c2 ........(2)The concentration of unknown solution can be determined by using equation (2). WhereA1= absorbance of known solutione 1= Molar absorptivity of known solutionc1= concentration of known solutionA2= absorbance of unknown solutione 2= Molar absorptivity of unknown solutionC2= concentration of unknown solution
For example concentration of unknown solution of nickel sulphate can be determined by using a known solution of same compound in colorimeter. This solution is known as standard solution and must have deep green color. A colorimeter consists of a red light from LED source which passes through solution and strike to photocell. Since slandered solution of nickel sulphate is deep in color, therefore more concentrated and absorbed more light than a solution of low concentration. This transmitted light is monitored by photocell. Measure the absorbance of a series of known solution prepared by using standard solution and plots a graph between absorbance and concentration of solutions. The graph shows a straight line with positive slop as according to Beer-Lambert law.Similarly measure the absorbance of unknown nickel sulphate solution with the colorimeter. Now we can locate the absorbance of this solution on the vertical axis of the graph and the corresponding concentration can be found on the horizontal axis.