Lesson Explainer: Molar Concentrations Chemistry

In this explainer, we will learn how to calculate the molar concentration of a solution from the solvent volume and mass or moles of the dissolved solute.

Solutions are frequently used in the chemistry laboratory. A solution is a homogeneous mixture consisting of one or more solutes dissolved in a solvent. The figure below shows four solutions of a red dye in water.

From visual inspection of the solutions, we can tell that the greatest amount of dye was dissolved in the leftmost solution, and the least amount of dye was dissolved in the rightmost solution. We can say that the leftmost solution is concentrated, containing a large amount of solute, and the rightmost solution is dilute, containing a small amount of solute. The terms concentrated and dilute are qualitative and do not indicate the actual amount solute present in the solution.

There are a variety of ways to express the quantitative amount of solute in solution. The concentration of a solution indicates the amount of a chemical constituent (solute, ions, or particles) with respect to the amount of solution. The amount of the constituent may be given as a mass, amount in moles, volume, or number of entities. It is very common for the amount of solution to be expressed as a mass or volume.

One specific type of concentration that is frequently calculated is the molar concentration, also called molarity or amount concentration. The molarity of a solution is the ratio of moles of solute per litre of solution. Molar concentration is sometimes simply referred to as concentration, although the term concentration can refer to other quantities.

Definition: Molarity

Molarity is the number of moles of solute per litre of solution. Molarity is also called molar concentration.

Molarity can be calculated by using the equation 𝑐=𝑛𝑉 or 𝑛=𝑐𝑉, where 𝑐 is the molar concentration, 𝑛 is the number of moles of solute, and 𝑉 is the total volume of the solution. The unit of 𝑛 must be moles and the unit of 𝑉 is typically litres. The unit of molarity will therefore be mol/L, molβ‹…Lβˆ’1, or simply M, read as β€œmolar.”

Equation: Molarity (c)

The following equation can be used to determine the molarity of a solution: 𝑛=𝑐𝑉, where 𝑛 is the amount of solute in moles, 𝑐 is the molar concentration typically in moles per litre (mol/L), and 𝑉 is the volume of the solution typically in litres.

Example 1: Calculating Molar Concentration from Moles and Volume

A 0.80-litre solution contains 0.40 moles of acetic acid. What is the molar concentration of acetic acid in this solution?

Answer

Molar concentration is a specific type of concentration defined as the number of moles of solute per litre of solution. It can be calculated using the molarity equation 𝑛=𝑐𝑉, where 𝑛 is the amount of solute in moles, 𝑐 is the molar concentration in moles per litre, and 𝑉 is the volume of the solution in litres.

We can substitute the amount of acetic acid in moles and the volume of the solution in litres into the equation, 0.40=𝑐⋅0.80,molL and rearrange to solve for the molar concentration: 0.400.80=𝑐0.5/=𝑐.molLmolL

The units mol/L and M can be used interchangeably when reporting the molar concentration. The molar concentration of acetic acid in the solution is 0.5 mol/L or 0.5 M.

It may be necessary when using the molarity equation to convert between mass and moles or between different volume units. We can convert between mass and moles using the following equation: 𝑛=π‘šπ‘€, where 𝑛 the amount in moles, π‘š is the mass in grams, and 𝑀 is the molar mass in grams per mole.

To convert between different volume units, it is important to recognize the following relationships: 1000=1=1=1000.mLLdmcm

Example 2: Calculating Concentration in Grams per Cubic Decimetre from Mass and Volume

A student dissolves 25 g of MgCl2 into water to produce a solution with a volume of 500 cm3. What is the concentration of the solution in moles per cubic decimetre? Give your answer to 2 decimal places. [Mg = 24 g/mol, Cl = 35.5 g/mol]

Answer

The concentration of a solution or the molar concentration is a ratio of the amount of solute in moles per volume of solution, and it can be calculated by using the following equation: 𝑐=𝑛𝑉, where 𝑐 is the molar concentration, 𝑛 is the amount of solute in moles, and 𝑉 is the volume of solution.

The amount of MgCl2, the solute, is given in grams but must be converted into moles. We can perform this conversion by using the following equation: 𝑛=π‘šπ‘€, where 𝑛 is the amount in moles, π‘š is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of MgCl2 can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+(2×𝑀)𝑀=24/+(2Γ—35.5/)𝑀=95/.(MgCl)(Mg)Cl(MgCl)(MgCl)222gmolgmolgmol

The mass given in the question and the molar mass can be substituted into the equation 𝑛=2595/ggmol to determine the amount of MgCl2 in moles to be 𝑛=0.26316….mol

The volume is given in cubic centimetres but the question asks for the concentration in moles per cubic decimetre. This means that the volume in cubic centimetres must be converted into cubic decimetres. There are 1β€Žβ€‰β€Ž000 cubic centimetres per cubic decimetre. We can multiply the volume in cubic centimetres by one cubic decimetres per 1β€Žβ€‰β€Ž000 cubic centimetres, 500Γ—11000,cmdmcm to determine the volume in cubic decimetres to be 0.5.dm

We can then substitute the amount of MgCl2 in moles and the volume of solution in cubic decimetres into the molarity equation, 𝑐=0.263160.5,moldm and calculate the concentration of the solution to be 0.52632… mol/dm3. Rounding our answer to 2 decimal places gives us a concentration of 0.53 mol/dm3.

Example 3: Calculating the Mass of Solute Needed to Prepare a Solution with a Desired Concentration and Volume

A student wants to prepare a 0.1 M solution of silver nitrate (AgNO3) in a volumetric flask that can contain 100 mL of water. How much silver nitrate does the student need to dissolve? Give your answer to 1 decimal place. [N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol]

Answer

The molar (M) is a unit of molar concentration and can be used in place of moles per litre (mol/L). The equation for molar concentration is 𝑛=𝑐𝑉, where 𝑛 is the amount of solute in moles, 𝑐 is the molar concentration typically in moles per litre, and 𝑉 is the volume of the solution typically in litres.

This question is asking us to determine the amount (𝑛) of silver nitrate, the solute. The molar concentration and volume of the solution were given. However, the volume must be converted into litres before it can be substituted into the equation.

There are 1β€Žβ€‰β€Ž000 millilitres in one litre. We can multiply the volume in millilitres by one litre per 1β€Žβ€‰β€Ž000 millilitres, 100Γ—11000=𝑉,mLLmL to determine the volume in litres to be 0.1.L

We can then substitute the concentration of the solution and the volume of the solution in litres into the concentration equation, 𝑛=0.1/β‹…0.1,molLL to determine the amount of silver nitrate needed to be 𝑛=0.01.mol

There is no way for the student to measure one mole of substance in the laboratory. As such, we will need to convert the amount of silver nitrate in moles to grams. The following equation can be used: 𝑛=π‘šπ‘€, where 𝑛 is the amount in moles, π‘š is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of silver nitrate can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀+(3×𝑀)𝑀=108/+14/+(3Γ—16/)𝑀=170/.(AgNO)(Ag)(N)(O)(AgNO)(AgNO)333gmolgmolgmolgmol

We can then substitute the amount in moles and the molar mass into the equation 0.01=π‘š170/molgmol and rearrange to solve for the mass of silver nitrate in grams: 170/Γ—0.01=π‘š1.7=π‘š.gmolmolg

The student will need to dissolve 1.7 grams of silver nitrate to produce 100 mL of a 0.1 M solution of silver nitrate.

Like molar concentration, we can also express the composition of a solution as a mass concentration where the 𝑛 in the equations 𝑐=𝑛𝑉 and 𝑛=𝑐𝑉 represents the amount of solute in grams instead of moles.

Example 4: Calculating Mass per Volume from Moles per Volume

What is the mass concentration of hydrogen chloride in 2.0 M hydrochloric acid? Give your answer to the nearest whole number. [H = 1 g/mol, Cl = 35.5 g/mol]

Answer

The molar (M) is the unit of molar concentration and can be used in place of moles per litre (mol/L). The equation for molar concentration can be expressed as 𝑐=𝑛𝑉, where 𝑛 is the amount of solute in moles, 𝑐 is the molar concentration in moles per litre, and 𝑉 is the volume of the solution in litres.

If the molar concentration of the solution is 2.0 mol/L, we can express this concentration as the following fraction: 𝑐=2.01.molL

This tells us that we can take 𝑛 to be 2.0 moles and 𝑉 to be 1 litre.

The question asks us to determine the mass concentration of hydrogen chloride (HCl) in this solution. The equation for mass concentration is the same as the equation for molar concentration except 𝑛 must be in grams.

We can convert the 2.0 moles of hydrogen chloride into grams of hydrogen chloride via the following equation: 𝑛=π‘šπ‘€, where 𝑛 is the amount in moles, π‘š is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of hydrogen chloride can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀𝑀=1/+35.5/𝑀=36.5/.(HCl)(H)(Cl)(HCl)(HCl)gmolgmolgmol

We can then substitute the amount of hydrogen chloride in moles and the molar mass into the equation 2.0=π‘š36.5/molgmol and rearrange to solve for the mass of hydrogen chloride: 36.5/Γ—2.0=π‘š73=π‘š.gmolmolg

The mass of hydrogen chloride in each litre of solution is 73 grams. We can substitute the amount in grams and the volume of 1 litre into the concentration equation, 𝑐=731,gL to determine the mass concentration of the solution to be 𝑐=73/.gL

The mass concentration of hydrogen chloride in 2.0 M hydrochloric acid is 73 g/L.

Changing the amount of solute, volume of solvent, or total volume of solution will change the concentration. Let us consider the solution of glucose (CHO6126) in water.

If more glucose is added to this solution, but the total volume remains constant, there will be more solute in the same volume as the original solution. Thus, the concentration will increase.

We can also see this effect by looking at the concentration equation: 𝑐=𝑛𝑉.

Adding more solute will increase the value of 𝑛. In an equation, increasing the value of the numerator will increase the resulting answer. Thus, increasing the amount of solute in a solution will increase the concentration.

Now let us consider what would happen if more water was added to the original solution. The number of glucose molecules will remain the same, but the molecules will spread out over a greater total volume. Thus, the concentration will have decreased.

We can also see this effect by looking at the concentration equation: 𝑐=𝑛𝑉.

Adding more solvent will increase the total volume of the solution, 𝑉. In an equation, increasing the value of the denominator will decrease the resulting answer. Thus, increasing the amount of solvent in a solution will decrease the concentration.

Example 5: Determining What Effect Increasing the Solute and Solvent Has on the Concentration of a Solution

A 1 M solution of CuSO4 is prepared by dissolving 159.5 g of CuSO4 in water to produce 1 litre of solution. [O = 16 g/mol, S = 32 g/mol, Cu = 63.5 g/mol]

  1. What would happen to the concentration if the amount of water was increased so that the total volume of solution doubled?
    1. The concentration would stay the same.
    2. The concentration would double.
    3. The concentration would halve.
  2. What would happen to the concentration if the amount of CuSO4 used to prepare the solution was doubled?
    1. The concentration would stay the same.
    2. The concentration would halve.
    3. The concentration would double.

Answer

Part 1

Water is the solvent in this solution. By increasing the solvent, the solute particles will be able to spread out further. Fewer solute particles in the same volume indicates that the concentration will decrease. We can quantitatively confirm this assumption.

The molar (M) is the unit of molar concentration and can be used in place of moles per litre (mol/L). The equation for molar concentration can be expressed as 𝑐=𝑛𝑉, where 𝑐 is the molar concentration typically in moles per litre, 𝑛 is the amount of solute in moles, and 𝑉 is the volume of the solution typically in litres. The amount of solute, CuSO4, is given in grams but must be converted into moles.

We can convert the 159.5 grams of copper(II) sulfate into moles via the following equation: 𝑛=π‘šπ‘€, where 𝑛 is the amount in moles, π‘š is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of copper(II) sulfate can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀+(4×𝑀)𝑀=63.5/+32/+(4Γ—16/)𝑀=159.5/.(CuSO)(Cu)(S)(O)(CuSO)(CuSO)444gmolgmolgmolgmol

We can then substitute the mass of copper(II) sulfate and the molar mass into the equation 𝑛=159.5159.5/ggmol and solve for the amount of copper(II) sulfate in moles: 𝑛=1.mol

The total volume of the solution has doubled from 1 litre to 2 litres. We can substitute the amount of solute in moles and the new volume of solution into the concentration equation, 𝑐=12,molL to determine the new concentration of the solution to be 𝑐=0.5/.molL

The original concentration was 1 M. By adding the additional water and doubling the total volume of the solution, the concentration has halved to 0.5 M. The correct answer is answer choice C.

Part 2

CuSO4 is the solute in this solution. By doubling the solute, there will be twice as many solute particles in the new solution as there were in the original solution. This corresponds to an increase in the concentration. We can quantitatively confirm this assumption.

We will once again use the equation 𝑐=𝑛𝑉.

The total volume of the solution will still be 1 litre. However, the mass of copper(II) sulfate dissolved in the solution has doubled to 319 grams. We can convert the 319 grams of copper(II) sulfate into moles via the following equation: 𝑛=π‘šπ‘€, where 𝑛 is the amount in moles, π‘š is the mass in grams, and 𝑀 is the molar mass in grams per mole. The molar mass of copper (II) sulfate is 159.5 g/mol as calculated in part 1.

We can substitute the mass of copper(II) sulfate and the molar mass into the equation 𝑛=319159.5/ggmol and solve for the amount of copper(II) sulfate in moles: 𝑛=2.mol

We can then substitute the new amount of solute in moles and the volume of solution into the concentration equation, 𝑐=21,molL to determine the new concentration of the solution to be 𝑐=2/.molL

The original concentration was 1 M. By doubling the solute, the concentration has doubled to 2 M. The correct answer is answer choice C.

The molar concentration is most commonly used in the chemistry laboratory. However, when we want to know the composition of a solution where temperature changes are involved, we cannot use molarity. Instead, we report the composition using molality (π‘š). The molality of a solution is the ratio of moles of solute per kilogram of solvent.

Definition: Molality (π‘š)

Molality is the number of moles of solute per kilogram of solvent.

Equation: Molality (π‘š)

The following equation can be used to determine the molality of a solution: π‘š=𝑛,m where π‘š is the molality, 𝑛 is the amount of solute in moles, and m is the mass of the solvent in kilograms.

As the molality is calculated by dividing the amount of solute in moles by the mass of the solvent in kilograms, the unit of molality will therefore be mol/kg, molβ‹…kgβˆ’1, or simply π‘š, read as β€œmolal.”

We can calculate the molality of a solution made by dissolving 28 g of potassium hydroxide (KOH) in 250 g of water. The molality of the solution can be calculated using π‘š=𝑛,m where π‘š is the molality, 𝑛 is the amount of solute in moles, and m is the mass of the solvent in kilograms.

We firstly need to calculate the number of moles of potassium hydroxide, using the equation 𝑛=π‘šπ‘€, where 𝑛 is the amount in moles, π‘š is the mass in grams, and 𝑀 is the molar mass in grams per mole.

The molar mass of potassium hydroxide can be calculated by summing the average molar masses of the constituent atoms: 𝑀=𝑀+𝑀+𝑀𝑀=39/+16/+1/𝑀=56/.(KOH)(K)(O)(H)(KOH)(KOH)gmolgmolgmolgmol

We can then substitute the mass of potassium hydroxide and the molar mass into the equation 𝑛=2856/ggmol and solve for the amount of potassium hydroxide in moles: 𝑛=0.5.mol

Next, we need the mass of solvent, remembering to convert the units into kilograms: 250Γ—11000=0.25.gkggkg

Now we can calculate the molality of the solution: π‘š=π‘›π‘š=0.50.25.mmolkg

The molality of the solution is therefore 2 mol/kg.

Key Points

  • The concentration of a solution is the ratio of the amount of a constituent per amount of solution.
  • The molar concentration is the ratio of the amount of solute in moles per volume of solution in litres.
  • The equation for molar concentration is 𝑛=𝑐𝑉, where 𝑛 is the amount of solute in moles, 𝑐 is the molar concentration, and 𝑉 is the volume of the solution in litres.
  • We can convert between volume units using the following relationships: 1000=1=1=1000.mLLdmcm
  • Increasing the amount of solute increases the concentration, while increasing the amount of solvent decreases the concentration.
  • Molality (π‘š) is defined as the number of moles of solute per kilogram of solvent, and it is used to represent the composition of a solution when the temperature is changing.

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