Chemistry
Molarity
Molarity is a measure of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution. It is commonly used in chemistry to quantify the amount of a substance in a given volume of solution. Molarity is calculated by dividing the number of moles of solute by the volume of the solution in liters.
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eBook - PDF- Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
- 2015(Publication Date)
- Openstax(Publisher)
The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution. A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration). Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution: M = mol solute L solution 150 Chapter 3 | Composition of Substances and Solutions This OpenStax book is available for free at http://cnx.org/content/col11760/1.9 Example 3.14 Calculating Molar Concentrations A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage? Solution Since the molar amount of solute and the volume of solution are both given, the Molarity can be calculated using the definition of Molarity. Per this definition, the solution volume must be converted from mL to L: M = mol solute L solution = 0.133 mol 355 mL × 1 L 1000 mL = 0.375 M Check Your Learning A teaspoon of table sugar contains about 0.01 mol sucrose.
eBook - ePub- Maria Glaucia Teixeira, Joel L. Zatz(Authors)
- 2017(Publication Date)
- Wiley(Publisher)
Molarity and molality are known pharmaceutically useful expressions of drug concentration. Both are based on the number of moles (or mols) of solute present in the solution but they are not the same expression. Please note that the spelling “mole” and “mol” are used interchangeably throughout this text.- Molarity (M) is the number of mols of solute per liter of solution.
- Molality (m) is the number of mols of solute per kilogram of solvent.
For example, a 1 M solution of acetic acid contains one mole of acetic acid per liter of solution. Because the molecular weight of acetic acid is 60, a 1 M solution of acetic acid contains 60 g of acetic acid per liter (60 g//L) or 60 mg/mL.In clinical practice it is traditional to only use “Molarity” as the majority of pharmaceutical solutions are frequently used in very dilute concentrations (density close to that of water) and are usually w/v systems.As we review mols and millimols, keep in mind the following:- One mol (or gram molecular weight) of a substance is defined as the formula weight for that substance, expressed in grams.
- A millimol (mmol) is the molecular weight expressed in milligrams. It is a measure of quantity, not concentration.
- A chemical equation relates quantities in terms of mols. By converting mols to grams, relationships in weight may be obtained.
- The number of mols of an ion or product is indicated by the coefficient (number), which precedes that species in the chemical equation.
9.1.1. Mols and Millimols
A mol (or mole) is the molecular weight of a substance in grams. The number of mols of a substance is calculated by dividing the number of grams of substance by the molecular weight in grams.Let us go through a couple of examples as a review of electrolytes' chemical equation and number of mols after dissociation.- Ex. 1. The chemical equation for the dissociation of calcium nitrate (Ca (NO3 )2 ) is
Therefore, one mol of calcium nitrate yields 1 mol of calcium ion and 2 mol of nitrate ion. We can also say that one mol of calcium nitrate, when dissociated, yields 3 ions. This concept will be important later when we calculate osMolarity of a solution (mosm/L).
- Ex. 2.
eBook - PDF- Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson(Authors)
- 2019(Publication Date)
- Openstax(Publisher)
A solution in which water is the solvent is called an aqueous solution. A solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration). Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution: EXAMPLE 3.14 Calculating Molar Concentrations A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage? 3.3 • Molarity 137 Solution Since the molar amount of solute and the volume of solution are both given, the Molarity can be calculated using the definition of Molarity. Per this definition, the solution volume must be converted from mL to L: Check Your Learning A teaspoon of table sugar contains about 0.01 mol sucrose. What is the Molarity of sucrose if a teaspoon of sugar has been dissolved in a cup of tea with a volume of 200 mL? Answer: 0.05 M EXAMPLE 3.15 Deriving Moles and Volumes from Molar Concentrations How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from Example 3.14? Solution Rearrange the definition of Molarity to isolate the quantity sought, moles of sugar, then substitute the value for Molarity derived in Example 3.14, 0.375 M: Check Your Learning What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example? Answer: 80 mL EXAMPLE 3.16 Calculating Molar Concentrations from the Mass of Solute Distilled white vinegar ( Figure 3.15) is a solution of acetic acid, CH 3 CO 2 H, in water.
eBook - PDFIntroductory Chemistry
An Active Learning Approach
- Mark Cracolice, Edward Peters, Mark Cracolice(Authors)
- 2020(Publication Date)
- Cengage Learning EMEA(Publisher)
Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 644 Chapter 16 Solutions Goal 9 Given two of the following, calcu- late the third: amount in moles of solute (or data from which it may be found), vol- ume of solution, Molarity. Molarity is the amount in moles of solute in one liter of solution. Units of Molarity are moles per liter, or mol/L. The symbol for Molarity is M. M K moles solute liter solution 5 mol L Goal 10 Given two of the following, cal- culate the third: moles of solute (or data from which it may be found), mass of sol- vent, molality. Molality is moles of solute in one kilogram of solvent. The symbol for molality is m (note that molality is lower case m and Molarity is upper case M). m K mol solute kg solvent Molality is used in situations where temperature independence is important. Neither the num- ber of solute particles nor the mass of the solvent varies with temperature. In contrast, molar- ity is temperature dependent because the volume of a solution varies with temperature. Goal 11 Given an equation for a neu- tralization reaction (or information from which it can be written), state the number of equivalents of acid or base per mole and calculate the equivalent mass of the acid or base. One equivalent of an acid is defined as that amount of acid that yields one mole of hydrogen ion in a specific reaction. One equivalent of a base is that amount of base that reacts with one equivalent of an acid. The equivalent mass of a substance is the mass in grams of the substance per equivalent.
- Chavan, U D(Authors)
- 2021(Publication Date)
- Daya Publishing House(Publisher)
For example, when 60 g of NaOH are dissolved in 1000 g of solvent, the solution contains 1.5 moles of solute in 1 kg of solvent. Therefore, the molality is 1.5. The factors needed to calculate molality are moles of solute and the mass of solvent in kilograms. The SI unit for molality is mol/kg. A solution with a molality of 3 mol/kg is often described as “3 molal” or “3 m.” The primary advantage of using molality to specify concentration is that unlike its volume, the mass of the solvent does not change with changes of temperature or pressure; molality remains constant under changing environment conditions. Molality, like mole fraction, is used in applications dealing with certain physical properties of solutions. As the molality changes, it affects the boiling point and freezing point (also known as the melting point) of the solution. A higher molality increases the boiling point and decreases the freezing point of the solution. As molality is a more accurate measure of solutes in solution in dynamic conditions, it is often used in comparing and determining colligative properties of solution. Molality is a property of solutions. If the solvent is reactive, and one needs to know the stoichiometry between the solvent and the solute, knowing the molality can be very important. The mass-based nature of molality implies that it can be readily converted into a mass ratio. As is Clear From its Name, Molality Involves Moles The molality of a solution is calculated by taking the moles of solute and dividing by the kilograms of solvent. This ebook is exclusively for this university only. Cannot be resold/distributed. Equivalent Concentration 105 Molality = kilograms of solvent moles of solute This is probably easiest to explain with examples. Example 1: Suppose we had 1.00 mole of sucrose (it’s about 342.3 grams) and proceeded to mix it into exactly 1.00 liter water.
eBook - PDFFoundations of Chemistry
An Introductory Course for Science Students
- Philippa B. Cranwell, Elizabeth M. Page(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
This section describes the various ways in which the concentration of a solu-tion can be expressed and the procedures and calculations involved in preparing solutions of a specific concentration by dilution. 3.4.1 Measuring and expressing concentrations When a substance is dissolved in a solvent such as water, a solution is formed. The concentration of the solution is a measure of how much of the substance is dissolved per unit volume of solvent (Figure 3.2). There are many different ways of expressing the concentration of a solution, but chemists prefer to work in mol dm -3 . One dm 3 (cubic decimetre) is the same volume as a litre or 1000 cm 3 . Therefore the unit of concentration is mol per dm 3 , written as mol dm -3 , and a concentration of 1 mol dm -3 is obtained when one mole of a substance is dissolved in 1 dm 3 of water. When a substance such as sugar dis-solves in water, a solution is formed. The substance that is being dissolved is called the solute, and the liquid it is dissolving in is called the solvent . For a reminder on units of volume, see Chapter 0 It is always safer to convert volumes to dm 3 to ensure the resulting con-centration is obtained in mol dm -3 . A volume in cm 3 or mL should be converted to dm 3 by multiplying by 10 -3 . A volume in litres can be chan-ged directly to dm 3 as 1 L is the same as 1 dm 3 . 86 Amount of Substance Concentration mol dm -3 = number of moles mol volume dm 3 or c = n V Using this equation, we can calculate concentrations of solutions but also work out the number of moles of a solute dissolved in a certain volume of solu-tion, if we know its concentration. Because solutions are normally made up in the laboratory by weighing out a certain amount of solid, it ’ s usually necessary to convert the mass of solute to an amount in moles to calculate the concentra-tion.
eBook - PDFChemistry
An Industry-Based Introduction with CD-ROM
- John Kenkel, Paul B. Kelter, David S. Hage(Authors)
- 2000(Publication Date)
- CRC Press(Publisher)
(10.10) defines Molarity. The number of mL given must first be converted to liters, and so 0.3000 L is used in the denominator. Example 10.5 What is the Molarity of a solution of NaOH that has 0.491 grams dissolved in 400.0 mL of solution? Solution 10.5 In order to use Eq. (10.10), the grams of solute must be converted to moles by dividing the grams by the formula weight (FW) as we did often in Chapters 8 and 9 [see Eq. (8.1) for cancellation of units]. Once again, the mL are converted to L. 10.6.3 Molality The molality , m, of a solution is the number of moles of solute dissolved per kilogram of solvent. (10.11) Notice that the denominator here is an amount of solvent and not of solution. Example 10.6 What is the molality of a solution of KCl if 0.722 grams of KCl are dissolved in 500.0 grams of water? Molarity moles of solute liters of solution -------------------------------------- Molarity 4.5 moles 0.3000 L ----------------------15 M Molarity grams/FW Liters of solution ----------------------------------------0.491 39.997 ( ) moles 0.4000 L --------------------------------------------------- 0.0307 M molality moles of solute kilograms of solvent ------------------------------------------------ 266 Chemistry: An Industry-Based Introduction with CD-ROM Solution 10.6 To calculate moles of KCl, the number of grams is divided by the formula weight. Also, the grams of water need to be converted to kilograms. 10.6.4 Parts per Million Parts per million , or ppm , is, as the name implies, the parts of solute per million parts of solution. A “part” can be a mass unit or a volume unit. Typically for solutions, the parts of solute is in milligrams of solute and the million parts of solution is in liters of solution. (10.12) Thus, if a solution of copper is labeled as 5 ppm, there are 5 milligrams of copper dissolved per liter of solution. The number of micrograms per mL, µ g/mL, is also ppm.
eBook - PDF- Allan Blackman, Steven E. Bottle, Siegbert Schmid, Mauro Mocerino, Uta Wille(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
Molarity (c) = amount of solute (mol) volume of solution (L) However, because solutions (usually) increase their volume when the temperature is raised, and vice versa, the Molarity of a solution changes as the temperature changes. For example, a saline solution of known Molarity prepared at 25 °C has a different Molarity when it is warmed to body temperature. Molality Molality is the preferred method of expressing solution composition in situations that involve colligative properties, as these properties are proportional to molality. We define the molal concentration or molality of a solution as the amount of solute per kilogram of solvent, and use the symbol b. molality (b) = amount of solute (mol) mass of solvent (kg) The molality of a solution is independent of temperature as mass does not depend on temperature. The Molarity and molality of aqueous solutions are numerically different but, as aqueous solutions become more and more dilute, the numerical value of the Molarity approaches that of the molality. WORKED EXAMPLE 10.6 Preparing a solution of a given molality What mass of NaCl would have to be dissolved in 500.0 g of water to prepare a solution of molality 0.150 mol kg -1 ? Analysis Rearranging the equation for molality given above, we obtain the following. amount of solute (mol) = molality (mol kg -1 ) × mass of solvent (kg) Knowing the amount of solute, NaCl, we then relate amount to mass through the molar mass of NaCl. Solution amount of NaCl = molality × mass of solvent = 0.150 mol kg -1 × 0.5000 kg = 0.0750 mol m = nM = 0.0750 mol × 58.44 g mol -1 = 4.38 g Therefore, 4.38 g of NaCl needs to be dissolved in 500.0 g of water to give a solution of molality 0.150 mol kg -1 . Is our answer reasonable? If we round the molar mass of NaCl to 60, 0.1 mol is 6 g and 0.2 mol is 12 g. We also notice that 0.150 mol kg -1 is halfway between 0.1 and 0.2 mol kg -1 . So, for 1 kg of water, we’d need halfway between 6 g (0.1 mol) and 12 g (0.2 mol) of NaCl.
- Premasis Sukul(Author)
- 2023(Publication Date)
Thus, it may be expressed by the following formula with a unit of ‘moles/kg’: Molality (m) = Number of moles of solute / Mass of solvent (kg) Equation 2.1 Thus, if a solution contains 5 moles of solute / kg of solvent, the solution is considered as 5 molal or 5m solution. Volumetric Analysis 19 2.3.1.2.1. Calculation of molality from a given mass Problem 1: Calculate the molality of a solution where 2 g NaCl is dissolved in 150 ml of water. Solution 1: First, mole of NaCl in 2 g NaCl is to be calculated. Moles of NaCl = Mass in g / Molecular weight = 2/58.4 = 0.03 mole Since density of water is 1 g/ml, 150 ml water is equivalent to 150 g or 0.150 kg water. Therefore, Molality = Moles of NaCl/Mass of water in kg = 0.03/0.150 m = 0.2 m 2.3.1.2.2. Calculating mass of a solute from given molality Problem 2: Calculate the mass of NaCl per kilogram of water in a 10 m aqueous solution. Solution 2: We know, Molality = Moles of solute/ mass of solvent in kg So, 10 = Moles of NaCl/1 Moles of NaCl = 10 Again, we know that Moles of solute = Mass in g/molar mass Or, 10 = Mass in g/58.4 Therefore, Mass of NaCl = 10 x 58.4 = 584 g Thus, 584 g NaCl is to be added to 1 kg water to make a 10 m NaCl solution. 2.3.1.3. Molarity (M) Molarity, also known as molar concentration, is defined as the total moles of a solute contained in a litre of a solvent. Thus, it may be expressed by the following formula with a unit of ‘moles/L’: 20 Principles of Analytical and Instrumental Techniques Molarity (M) = Number of moles of solute/volume of solvent (L) Equation 2.2. So, we may write 1M (molar solution) = 1 mol/ L = 1 mmol/mL = 1 µmol/µL 1mM (millimolar solution) = 1 mmol/L = 1 µmol/mL Thus, if a solution contains 5 moles of solute/L of solvent, the solution is considered as 5 molar or 5M solution. Therefore, the 5M NaCl solution contains 5x58.46 g of NaCl in I L distilled or deionized water.
- James N. Jensen(Author)
- 2022(Publication Date)
- Wiley(Publisher)
2.3 | Molar Concentration Units 15 In molar units, the constant relating moles to numbers is Avogadro’s number (6.022×10 23 atoms, molecules, or ions per mole). The molar system is the only set of concentration units with the same constant relating the concentration units to the number of atoms, molecules, or ions for all chemical species. Another way to express this idea is to take the approach of Worked Example 2.1. For any set of concentration units, you can write something L number L X . Molar units (where: something = moles) is the only set of concentration units where X is constant for all chemical species. To summarize, molar units are the only concentration units with the same constant relating the concentration units to the number of atoms, molecules, or ions for all chemical species. Thus, molar units are especially useful in equilibrium calculations, where combining ratios are critical. You should always use molar units in equilibrium calculations for the concentrations of dissolved species. 2.3.3 Molarity As will be discussed in the Historical Note, Avogadro’s number was set by defining the mass of 1 mole of the 12 C isotope of carbon atoms equal to exactly 12 grams. (The 12 C isotope of carbon has 6 neutrons in addition to 6 protons and 6 electrons.) With Avo- gadro’s number fixed, you can use the combining ratios of the elements to calculate Units Conversion How do you convert the concentration units of milligrams per liter (mg/L) to the concentration units of moles per liter (mol/L)? Solution To answer this question, you do not required values. (In fact, you do not even need to know what the words milligrams and moles mean.) Simply treat the units as terms in an equation: mg L mol L X Solving for X: X mol L L mg mol mg Thus, to convert from mg/L to mol/L, multiply by the number of moles per milligram.
eBook - PDF- William R. Robinson, Edward J. Neth, Paul Flowers, Klaus Theopold, Richard Langley(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
The empirical formula mass of a covalent compound may be compared to the compound’s molecular or molar mass to derive a molecular formula. 6.3 Molarity Solutions are homogeneous mixtures. Many solutions contain one component, called the solvent, in which other components, called solutes, are dissolved. An aqueous solution is one for which the solvent is water. The concentration of a solution is a measure of the relative amount of solute in a given amount of solution. Concentrations may be measured using various units, with one very useful unit being Molarity, defined as the number of moles of solute per liter of solution. The solute concentration of a solution may be decreased by adding solvent, a process referred to as dilution. The dilution equation is a simple relation between concentrations and volumes of a solution before and after dilution. 6.4 Other Units for Solution Concentrations In addition to Molarity, a number of other solution concentration units are used in various applications. Percentage concentrations based on the solution components’ masses, volumes, or both are useful for expressing relatively high concentrations, whereas lower concentrations are conveniently expressed using ppm or ppb units. These units are popular in environmental, medical, and other fields where mole-based units such as Molarity are not as commonly used. Exercises 6.1 Formula Mass and the Mole Concept 1. What is the total mass (amu) of carbon in each of the following molecules? (a) CH 4 (b) CHCl 3 (c) C 12 H 10 O 6 (d) CH 3 CH 2 CH 2 CH 2 CH 3 Chapter 6 | Composition of Substances and Solutions 333 2. What is the total mass of hydrogen in each of the molecules? (a) CH 4 (b) CHCl 3 (c) C 12 H 10 O 6 (d) CH 3 CH 2 CH 2 CH 2 CH 3 3. Calculate the molecular or formula mass of each of the following: (a) P 4 (b) H 2 O (c) Ca(NO 3 ) 2 (d) CH 3 CO 2 H (acetic acid) (e) C 12 H 22 O 11 (sucrose, cane sugar).
eBook - PDF- Leo J. Malone, Theodore O. Dolter(Authors)
- 2012(Publication Date)
- Wiley(Publisher)
p. 400 12-2.1 Solubility is maximum amount of a solute that dissolves at a specific temperature. p. 401 12-2.1 Depending on the amount of solute dissolved and its solubility, the solution may be unsaturated, saturated, or, in certain circumstances, supersaturated. p. 401 12-2.1 The concentration of a solute refers to the amount present in a certain amount of solvent or solution. p. 401 12-2.2 Recrystallization takes advantage of the difference in solubility of a substance at dif- ferent temperatures. p. 402 416 CHAPTER 12 Aqueous Solutions 12-2.4 According to Henry’s law, the solubility of gases in a liquid relates to the pressure above the liquid. p. 403 12-3 Concentration may be measured as percent by mass. p. 404 12-3.1 Parts per million (ppm) or parts per billion (ppb) are used for small concentrations. p. 405 12-4.1 Molarity (M) is a concentration unit that emphasizes the volume of the solution. p. 407 12-5.2 During a titration, an indicator can be used to determine the end point of the experiment. p. 414 Concentration Units Concentration Unit Name Relationship of solute Use g solute 100 g solvent ——— Mass of solvent Solubility tables g solute g solution * 100, Percent by mass Mass of solution High concentrations (above 0.01%) g solute g solution * 10 6 ppm Parts per million (ppm) Mass of solution Low concentrations (710 -4 %) g solute g solution * 10 9 ppb Parts per billion ppb Mass of solution Extremely low concentrations (710 -7 %) mole solute L solution Molarity (M) Volume of solution Measuring molar amount with volume and in stoichiometry problems SUMMARY CHART OBJECTIVES 12-6 Explain the differences between nonelectrolytes, strong electrolytes, and weak electrolytes. 12-7 Calculate the boiling and melting points of aqueous solutions of electrolytes and nonelectrolytes. SETTING A GOAL ■ n You will learn how the physical properties of aqueous solutions differ from those of pure water.
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