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Molarity
Molarity is defined as the moles of a solute per volume of total solution.
Learning Objectives

Determine the amount or volume needed to complete a dilution procedure

Determine the molarity of a solution as well as calculate the volume or amount required of a solution using molarity
Key Points
 Molarity (M) indicates the number of moles of solute per liter of solution (moles/Liter) and is the unit used to measure the concentration of a solution.
 Molarity can be used to calculate the volume of solvent or the amount of solute.
 The relationship between two solutions with the same dilutions can be represented by the formula c_{1}V_{1} = c_{2}V_{2}, where c is concentration and V is volume.
Terms

dilution
The process by which a solution is made less concentrated via addition of more solvent.

concentration
The proportion of a substance in a mixture.

SI unit
The modern form of the metric system used extensively in the sciences (abbreviated SI from French: Système International d'Unités).

molarity
The concentration of a substance in solution, expressed as the number moles of solute per liter of solution.
Full Text
In chemistry, concentration of a solution is often measured in molarity (M), which is the number of moles per liter of solution. This molar concentration (c_{i}) is calculated by dividing the moles of solute (n_{i }) by the total volume (V):
The SI unit for molar concentration is mol/m^{3}. However, mol/L is a more common unit for molarity. A solution that contains 1 mole of solute per 1 liter of solution (1 mol/L) is called "one Molar" or 1 M. The unit mol/L can be converted to mol/m^{3} using the following equation:
1 mol/L = 1 mol/dm^{3} = 1 mol dm^{−3} = 1 M = 1000 mol/m^{3}
Calculating Molarity
To calculate the molarity of a solution, the number of moles of solute must be divided by the total liters of solution produced. If the amount of solute is given in grams, we must first calculate the number of moles of solute using the solute's molar mass, then calculate the molarity using the number of moles and total volume.
Calculating Molarity Given Moles and Volume
For example, if there are 10.0 grams of NaCl (the solute) dissolved in enough water (the solvent) to produce 2.0 L of solution, what is the molarity of this solution?
First, we must convert the mass of NaCl in grams into moles. We do this by dividing by the molecular weight of NaCl (58.4 g/mole).
Then, we divide the number of moles by the total solution volume to get concentration.
The NaCl solution is a 0.1 M solution.
Calculating Moles Given Molarity
To calculate the number of moles in a solution given the molarity, we multiply the molarity by total volume of the solution in liters.
How many moles of potassium chloride (KCl) are in 4.0 L of a 0.65 M solution?
There are 2.6 moles of KCl in a 0.65 M solution that occupies 4.0 L.
Calculating Volume Given Molarity and Moles
We can also calculate the volume required to meet a specific mass in grams given the molarity of the solution. This is useful with particular solutes that cannot be easily massed with a balance. For example, diborane (B_{2}H_{6}) is a useful reactant in organic synthesis, but is also highly toxic and flammable. Diborane is safer to use and transport if dissolved in tetrahydrofuran (THF).
How many milliliters of a 3.0 M solution of BH3THF are required to receive 4.0 g of BH3?
First we must convert grams of BH_{3} to moles by dividing the mass by the molecular weight.
Once we know we need to achieve 0.29 moles of BH_{3,} we can use this and the given molarity (3.0 M) to calculate the volume needed to reach 4.0 g.
Now that we know that there are 4.0 g of BH_{3 }present in 0.1 L, we know that we need 100 mL of solution to obtain 4.0 g of BH_{3}.
Dilution
Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. This relationship is represented by the equation c_{1}V_{1} = c_{2}V_{2} , where c_{1} and c_{2} are the initial and final concentrations, and V_{1} and V_{2} are the initial and final volumes of the solution.
Example 1
A scientist has a 5.0 M solution of hydrochloric acid (HCl) and his new experiment requires 150.0 mL of 2.0 M HCl. How much water and how much 5.0 M HCl should the scientist use to make 150.0 mL of 2.0 M HCl?
c_{1}V_{1} = c_{2}V_{2}
c_{1} and V_{1 }are the concentration and the volume of the starting solution, which is the 5.0 M HCl. c_{2} and V_{2 }are the concentration and the volume of the desired solution, or 150.0 mL of the 2.0 M HCl solution. The volume does not need to be converted to liters yet because both sides of the equation use mL. Therefore:
V_{1} = 60.0 mL of 5.0 M HCl
If 60.0 mL of 5.0 M HCl is used to make the desired solution, the amount of water needed to properly dilute the solution to the correct molarity and volume can be calculated:
150.0 mL  60.0 mL = 90.0 mL
In order for the scientist to make 150.0 mL of 2.0 M HCl, he will need 60.0 mL of 5.0 M HCl and 90.0 mL of water.
Example 2
Water was added to 25 mL of a stock solution of 5.0 M HBr until the total volume of the solution was 2.5 L. What is the molarity of the new solution?
We are given the following: c_{1}= 5.o M, V_{1}= 0.025 L, V_{2}= 2.50 L. We are asked to find c_{2}, which is the molarity of the diluted solution.
(5.0 M)(0.025 L) = c_{2} (2.50 L)
Notice that all of the units for volume have been converted to liters. We calculate that we will have a 0.05 M solution, which is consistent with our expectations considering we diluted 25 mL of a concentrated solution to 2500 mL.
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Key Term Reference
 acid
 Appears in these related concepts: Balancing Redox Equations, Basic and Amphoteric Hydroxides, and Calculating Percent Dissociation
 chemistry
 Appears in these related concepts: Bond Polarity, Coloring Agents, and The Study of Chemistry
 liter
 Appears in these related concepts: Base Dissociation Constant, Effect of a Common Ion on Solubility, and Molar Solubility and Relative Solubility
 molar mass
 Appears in these related concepts: Converting between Mass and Number of Moles, MasstoMole Conversions, and Molar Mass of Compounds
 mole
 Appears in these related concepts: Avogadro's Number and the Mole, Gas Exchange across the Alveoli, and Concept of Osmolality and Milliequivalent
 reactant
 Appears in these related concepts: The Law of Conservation of Mass, Writing Chemical Equations, and Chemical Reactions and Molecules
 solute
 Appears in these related concepts: Osmotic Pressure, Solubility, and Osmosis
 solution
 Appears in these related concepts: Electrolyte and Nonelectrolyte Solutions, Using Molarity in Calculations of Solutions, and Turning Your Claim Into a Thesis Statement
 solvent
 Appears in these related concepts: Empirical Formulas, Solution Stoichiometry, and Properties of Sulfur
 toxic
 Appears in these related concepts: The Halogens (Group 17), OddElectron Molecules, and Occupational Safety and Health Act
 volume
 Appears in these related concepts: Volume and Density, Shape and Volume, and Line
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Source: Boundless. “Molarity.” Boundless Chemistry. Boundless, 21 Jul. 2015. Retrieved 14 Feb. 2016 from https://www.boundless.com/chemistry/textbooks/boundlesschemistrytextbook/solutions12/concentrationunits93/molarity4004501/