Molarity

Molarity

Introduction

When chemists prepare solutions in the laboratory, they need a precise way to describe how much dissolved material is present in a given volume of liquid. This property is called concentration. Of all the ways to express concentration, molarity is by far the most commonly used in general chemistry, analytical chemistry, and the biological sciences. It is the concentration unit you will encounter most often in reaction calculations, titrations, and solution preparation.

This lecture defines molarity, explains what it means physically, and develops the core skills of:

• Calculating molarity from measured quantities

• Finding the number of moles in a given volume

• Preparing a solution of specified concentration

• Performing dilution calculations

Figure 1: Solutions in the Laboratory

Explore the process of preparing solutions in the laboratory, highlighting the use of volumetric flasks and different methods for measuring concentration, such as molarity, spectrophotometry, and titration.

Solute, Solvent, and Solution

Before defining molarity, it is important to understand the terminology used for solutions.

A solution is a homogeneous mixture of two or more substances.

The solute is the substance being dissolved.

The solvent is the dissolving medium.

For example, in an aqueous solution of sodium chloride:

This table displays the components of a saline solution, including the solute NaCl and the solvent H₂O, to illustrate how they combine to form NaCl(aq).

Typically:

• The solute is present in the smaller amount.

• The solvent is present in the larger amount.

Understanding these terms is important because different concentration units use different denominators.

For example:

Table displaying various concentration measures including molarity, molality, and weight percent to facilitate understanding of solution composition.

Key Point

The solute is what is dissolved, the solvent is what dissolves it, and the solution is the homogeneous mixture of both.

Figure 2: Solute, Solvent, and Solution

Explore the process of forming a solution, where salt (solute) dissolves in water (solvent) to create saltwater, demonstrating how water molecules separate and disperse ions evenly.

Defining Molarity

Molarity is defined as:

M=moles of soluteliters of solutionM=\frac{\text{moles of solute}}{\text{liters of solution}}

Units:

mol/L\mathrm{mol/L}

or simply:

M\mathrm{M}

A 1.0 M sodium chloride solution contains:

1.0 mol NaCl1.0\ \text{mol NaCl}

dissolved in enough water to produce:

1.0 L solution1.0\ \text{L solution}

Notice the wording carefully.

The solution contains:

1.0 L solution1.0\ \text{L solution}

not:

1.0 L water1.0\ \text{L water}

This distinction becomes extremely important when preparing solutions in the laboratory.

Key Point

Molarity equals moles of solute divided by liters of total solution, not liters of solvent.

Figure 3: Visualizing a 1.0 M Solution

Understand the concept of a 1.0 Molar (M) solution by visualizing 1 mole of sodium chloride (NaCl) dissolved in water to make exactly 1.000 liter of solution, highlighting the relationship between moles, mass, and volume.

The Molarity Relationship

The molarity equation can be rearranged to solve for any of the three variables.

Starting with:

M=molLM=\frac{\text{mol}}{\text{L}}

Solve for Moles

mol=M×L\text{mol}=M \times L

Solve for Volume

L=molML=\frac{\text{mol}}{M}

These three equations form the foundation of nearly all solution calculations.

Figure 4: Molarity Equation Triangle

Explore the molarity equation triangle, illustrating the relationships among molarity (M), moles of solute (n), and liters of solution (L), with a practical example of dissolving NaCl.

Worked Example: Calculating Molarity

Problem

A solution contains:

0.350 mol NaCl0.350\ \text{mol NaCl}

in:

0.500 L solution0.500\ \text{L solution}

Find the molarity.

Step 1: Use the molarity equation

M=molLM=\frac{\text{mol}}{\text{L}}

Step 2: Substitute values

M=0.3500.500M=\frac{0.350}{0.500}

Step 3: Calculate

M=0.700M=0.700

Answer

0.700 M0.700\ \mathrm{M}

Key Point

Molarity is determined by dividing moles of solute by liters of solution.

Worked Example: Finding Moles from Molarity

Problem

How many moles of NaCl are contained in:

0.500 L0.500\ \text{L}

of a:

0.700 M0.700\ \mathrm{M}

solution?

Step 1: Use the rearranged equation

mol=M×L\text{mol}=M \times L

Step 2: Substitute values

=(0.700)(0.500)=(0.700)(0.500)

Step 3: Calculate

=0.350 mol=0.350\ \text{mol}

Answer

0.350 mol NaCl0.350\ \text{mol NaCl}

Figure 5: Finding Moles from Molarity

This guide illustrates the steps to calculate moles of NaCl in a solution using the molarity equation, demonstrating the conversion from milliliters to liters for accurate results.

Worked Example: Converting Moles to Grams

Suppose we wish to determine how many grams correspond to:

0.350 mol NaCl0.350\ \text{mol NaCl}

Step 1: Use the molar mass

58.44 g/mol58.44\ \mathrm{g/mol}

Step 2: Convert moles to grams

0.350 mol×58.44 gmol0.350\ \text{mol}\times58.44\ \frac{\text{g}}{\text{mol}}

Step 3: Calculate

=20.45 g=20.45\ \text{g}

Answer

20.4 g NaCl20.4\ \text{g NaCl}

(rounded appropriately)

Key Point

Moles multiplied by molar mass gives grams.

Figure 6: Moles-to-Grams Conversion

A step-by-step guide to converting moles of a substance to grams using the molar mass, illustrated with the example of calculating the grams of NaCl in 0.250 moles.

Preparing a Solution of Known Molarity

In the laboratory, preparing a solution of a desired molarity requires two major steps.

Step 1: Calculate the Required Solute Mass

Determine how many moles are needed.

Convert those moles into grams.

Step 2: Prepare the Solution

1. Weigh the required mass of solute.

2. Dissolve it in a small amount of solvent.

3. Transfer the solution quantitatively into a volumetric flask.

4. Add solvent until the calibration mark is reached.

Important Laboratory Rule

Never:

• Fill the flask to the mark with pure solvent first

• Then add the solute

The final volume must include both solute and solvent.

Volumes are not necessarily additive.

Key Point

Always dissolve the solute first and then dilute to the final volume in a calibrated volumetric flask.

Figure 7: Preparing a Solution in a Volumetric Flask

Step-by-step guide to preparing a solution in a volumetric flask, ensuring precise volume and concentration for accurate scientific results.

Dilution: M₁V₁ = M₂V₂

Dilution calculations are among the most common concentration calculations.

When a solution is diluted:

• Solvent is added

• The volume increases

• The concentration decreases

However, the number of moles of solute remains unchanged.

This relationship gives the dilution equation:

M1V1=M2V2M_1V_1=M_2V_2

where:

• M1M_1 Molarity of the solution you wish to make

• V1V_1 Volume of the solution you wish to make

• M2M_2 Molarity of the stock solution

• V2V_2 Volume of stock solution needed (equal in moles to the solution you wish to make)

Figure 8: Dilution Process

Illustration of the dilution process, showing step-by-step instructions for preparing a less concentrated solution from a stock solution using a pipette and volumetric flask.

Worked Example: Dilution

Problem

How many milliliters of a:

2.0 M2.0\ \mathrm{M}

NaCl stock solution are required to prepare:

25.0 mL25.0\ \text{mL}

of:

0.75 M0.75\ \mathrm{M}

NaCl?

Step 1: Use the dilution equation

M1V1=M2V2M_1V_1=M_2V_2

Step 2: Solve for V1V_1V1​

V1=M2V2M1V_1=\frac{M_2V_2}{M_1}

Step 3: Substitute values

=(0.75)(0.025)2.0=\frac{(0.75)(0.025)}{2.0}

Step 4: Calculate

=0.0094 L=0.0094\ \text{L}

Step 5: Convert to mL

0.0094 L×1000=9.4 mL0.0094\ \text{L}\times1000=9.4\ \text{mL}

Answer

9.4 mL9.4\ \text{mL}

Measure out 9.4 mL of stock solution and dilute to a final volume of 25.0 mL.

Key Point

Dilution conserves moles of solute. Measure the required volume of stock solution and then dilute to the desired final volume.

Summary

Molarity is the most commonly used concentration unit in chemistry and is defined as moles of solute per liter of solution. Understanding the distinction between solute, solvent, and solution is essential for correctly applying concentration units. The molarity equation can be rearranged to solve for moles, volume, or concentration, and it serves as the basis for many laboratory calculations. Preparing solutions requires careful volumetric technique, while dilution calculations rely on the conservation of moles expressed through the equation M1V1=M2V2M_1V_1 = M_2V_2​.

Key Points

• Molarity is defined as moles of solute per liter of solution.

• The symbol for molarity is M.

• Molarity uses total solution volume, not solvent volume.

• The relationship M=molLM = \frac{\text{mol}}{L} can be rearranged to solve for any variable.

• Moles can be converted to grams using molar mass.

• Solutions should be prepared using calibrated volumetric flasks.

• Dissolve solute before diluting to the final volume.

• Volumes are not necessarily additive.

• Dilution conserves moles of solute.

• The dilution equation is M1V1=M2V2M_1V_1 = M_2V_2.