## What is molarity?

When working with solutions in a laboratory setting, we often express concentrations using one of the following:

- Volume/volume percent | (v/v)%
- Mass/volume percent | (m/v)%
- Parts per million | ppm
- Molarity | M or mol/L

Following on the previous article in this series, we will now turn our attention on one of the most common ways of expressing the concentration – using molarity.

**Molarity**, sometimes referred to as molar concentration, is defined as the concentration of a substance in solution, expressed as the number moles of solute per litre of solution. Molar (symbol M or mol/L) is the unit used to express molar concentrations. A solution with a concentration of 1 M has 1 mole of the substance dissolved per 1 litre of a solvent.

- 1 M of NaCl in water = 1 mole of NaCl per 1 litre of water.
- 1 mol/L of NaCl in water = 1 mole of NaCl per 1 litre of water.
- 0.05 M of oxalic acid in acetone = 0.05 moles of oxalic acid per 1 litre of acetone
- 0.05 mol/L of oxalic acid in acetone = 0.05 moles of oxalic acid per 1 litre of acetone
- 0.03 M of aspirin in ethanol = 0.03 moles of aspirin per 1 litre of ethanol.
- 0.03 mol/L of aspirin in ethanol = 0.03 moles of aspirin per 1 litre of ethanol.

## How to do molarity calculations:

Typically in a laboratory setting such as ones in an academic setting, it isn’t uncommon to work with volumes of solvent less than a litre for health and safety reasons and to curtail costs. In such cases, known amounts of chemical substance are dissolved in volumes lower than one litre and the concentration is generally expressed in molarity (M or mol/L). Note that some use M and mol/L interchangeably.

We will now go through some examples to try and help you understand how to calculate the molarity and how to do calculations to do with molarity. You will need a periodic table of elements for this.

## Problem 1:

*“0.02 mol of D-Xylose was dissolved in 160 ml of water. Express the concentration in mol/L.”*

The relationship between molarity, moles and volume are shown below.

where M = molarity (M or mol/L), n = number of moles (mol) and v = volume (L).

**Step 1:**

Determine which formula to use.

We were given the number of moles of the substance and the volume of solvent used. We do not know the concentration in molarity. In order to solve Problem 1, we will need to use Formula 1.

**Step 2:**

Conversion of units.

The number of moles of D-Xylose is already expressed in mol so we do not need to do any conversion here. On the other hand, the volume of the solvent is expressed in ml so we will need to convert the volume from ml to l.

160 ml of water = 0.160 L of water.

**Step 3:**

Use Formula 1.

Now that we are using the correct units, we may now use Formula 1 to determine the concentration in molarity.

**Answer to Problem 1:**

The concentration of the aqueous D-xylose solution is 0.125 mol/L or 0.125 M.

## Problem 2:

*“An unknown amount (in grams) of the antifungal miconazole (structure and chemical formula shown below) was dissolved in 50 ml of solvent to form a solution with a concentration of 0.001 M. Determine the mass of miconazole used to form the aforementioned solution in milligrams.”*

C_{18}H_{14}C_{l4}N_{2O}

**Step 1:**

Determine which formula(s) to use.

We were given the volume and the concentration. We do not know the number of moles. For this reason, we will need to use Formula 2 to determine the number of moles of miconazole used. Using the relationship between mass, moles and molecular weight, we can convert the number of moles to mass later (see previous article).

**Step 2:**

Conversion of units.

The concentration is already expressed in M so we do not need to do any conversion here. On the other hand, the volume of the solvent is expressed in ml so we will need to convert the volume from ml to l.

50 ml of solvent = 0.050 l of solvent.

**Step 3:**

Use Formula 2.

Now that we are using the correct units, we may now use Formula 1 to determine the number of moles.

**Step 4:**

Convert moles to mass

Recall from the previous article that the relationship between moles and mass is as follows:

Using a periodic table and the provided chemical formula of miconazole, the molecular weight of miconazole is determined as 416.13 g/mol. Now that we know both the number of moles of miconazole and its molecular weight, we can now use the formula above to determine the mass of miconazole used to form the 0.001 M solution.

**Step 5:**

Convert grams to milligrams.

We were asked to express the mass in mg.

0.0208 g = 20.8 mg

**Answer to Problem 2:**

20.8 mg of miconazole was dissolved in 50 ml of solvent to form a solution with a concentration of 0.001 M.

## Problem 3:

*“A trainee prepared a solution of cis-(−)-carveol (152.24 g/mol), a component of spearmint oil, for polarimetry studies. Given that 2200 mg of cis-(−)-carveol was dissolved in 20 ml of solvent to form the aforementioned solution, express the concentration of the solution prepared by the trainee in molarity (M).”*

**Step 1:**

Determine which formula(s) to use.

Recall the relationship between mass, moles and molecular weight. We were given the volume of solvent and the mass and molecular weight of cis-(−)-carveol. We do not know the concentration in molarity. For this reason, we will need to use Formula 1. However, we are given the mass rather than the number of moles. For this reason, we will need to convert mass to moles using one of the formulas shown in the previous article.

**Step 2:**

Conversion of units.

All the formulas that we need to solve this problem requires the mass and the volume in grams (g) in litres (l), respectively.

2200 mg of cis-(−)-carveol = 2.2 g of cis-(−)-carveol.

20 ml of solvent = 0.020 l of solvent.

**Step 3:**

Use the formulas.

**Converting mass to moles:**

**Use Formula 1:**

**Answer to Problem 3:**

Using 2200 mg of substance dissolved in 20 ml of solvent, the trainee prepared a solution of cis-(−)-carveol with a concentration of 0.725 mol/L for polarimetry studies.

## Problem 4:

*“Suppose you were tasked to make up 50 ml of 25 mM aqueous solution of KBr. How much KBr (in grams) is needed to form 50 ml of 25 mM aqueous solution of KBr?”*

**Step 1:**

Determine which formula(s) to use.

We were given the volume of solvent and the concentration of the solution. We do not know the amount of KBr required. For this reason, we will need to use Formula 2. Using the relationship between moles, molecular weight and mass, we can determine the amount of KBr we need to form 50 ml of 25 mM aqueous solution of KBr.

We weren’t provided the molecular weight of KBr so we will need to consult a periodic table to determine its molecular weight by using the sum of the atomic weights of K and Br (works out as 119.0 g/mol). Aqueous means that the solvent is water.

**Step 2:**

Conversion of units.

Conversion from millilitre to litre:

50 ml = 0.05 l

Conversion from millimolar to molar:

25 mM = 0.025 M

**Step 3:**

Use the formulas

Conversion from millilitre to litre:

50 ml = 0.05 l

Conversion from millimolar to molar:

25 mM = 0.025 M

Converting moles to mass:

**Answer to Problem 4:**

0.14875 g of KBr needs to be dissolved in 50 ml of water to form a 50 ml 25 mM aqueous solution of KBr.

Attempt to do all four problems on your own and check your answers.