# Moles and Solutions – A Level Chemistry Revision

## Welcome back to Beyond’s Science Blog! This A Level Chemistry post explores Moles and Solutions.

You can also subscribe to Beyond Secondary Resources for access to thousands of worksheets and revision tools. Our site was created with teachers in mind and includes lots of teacher instructions, however, it also contains content for students that will be particularly useful when revising! You can sign up for a free account here and take a look around at our free resources before you subscribe too.

A solution is formed when a solute is dissolved in a solvent. For example, when sodium chloride, NaCl(s), dissolves in water, sodium chloride solution, NaCl(aq), is formed.

The concentration of a solution describes the amount of solute in a certain volume of solvent and can be calculated using the equation:

$\text{concentration (mol dm}^{-3}\text{)} = \frac{\text{amount of solute (mol)}}{\text{volume of solvent (dm}^3\text{)}}$

For example:

1. Calculate the concentration of a solution that contains 0.5 moles of potassium iodide dissolved in 1.0 dm3 of water.

concentration = $\frac{0.5}{1.0}$ = 0.5 mol dm-3

2. Calculate the concentration of a solution containing 1.0 mole of sodium hydroxide dissolved in 10 dm3 of ethanol.

concentration = $\frac{1}{10}$ = 0.1 mol dm-3

The above equation can be rearranged to calculate the number of moles of a solute in a solution of known volume and concentration as follows:

$\text{number of moles (mol)} = \text{volume (dm}^3\text{)}\,\times\,\text{concentration (mol dm}^{-3}\text{)}$

For example:

1. Calculate the number of moles of copper chloride in 1.0 dm3 of a 0.25 mol dm-3 copper chloride solution.

number of moles = 1.0 × 0.25 = 0.25 mol

2. Calculate the number of moles of hydrochloric acid in 0.5 dm3 of a 1.5 mol dm-3 hydrochloric acid solution.

number of moles = 0.5 × 1.5 = 0.75 mol

It is important to read information relating to solutions carefully. In practical situations, the volumes of liquids are often recorded in cubic centimetres (cm3) rather than cubic decimetres (dm3) and amounts of solutes are often recorded in grams (g) rather than moles (mol). Therefore, it is often necessary to convert the units of given volumes, amounts or concentrations before using them in your calculations.

$\text{number of moles (mol)} = \frac{\text{mass of substance (g)}}{\text{A}_{\text{r}}\text{ or M}_{\text{r}}\text{ of substance (g mol}^{-1}\text{)}}$

Example Question 1

Calculate the concentration, in mol dm-3, of a solution that contains 37.3 g of potassium chloride (KCl) in 500 cm3 of water.

Mr of KCl = (39.1 x 1) + (35.5 x 1) = 74.6

moles of KCl = $\frac{37.3}{74.6}$ = 0.5

volume = $\frac{500\,\text{cm}^3}{1000}$ = 0.5dm3

concentration = $\frac{0.5}{0.5}$ = 1 mol dm-3

Example Question 2

Calculate the mass, in grams, of iron(II) sulfate (FeSO4) in 50 cm3 of 0.25 mol dm-3 iron(II) sulfate solution.

volume = $\frac{50\,\text{cm}^3}{1000}$ =0.05 dm3

number of moles of FeSO4 in solution = 0.05 x 0.25 = 0.0125

Mr of FeSO4 = (55.8 x 1) + (32.1 x 1) + (16.0 x 4) = 151.9

mass = 0.0125 x 151.9 = 1.89875 g

mass = 1.9 g (2s.f.)

Example Question 3

Calculate the concentration of a 20 g L-1 (grams per litre) solution of hydrogen peroxide (H2O2) in mol dm-3.

Mr of H2O2 = (1.0 x 2) + (16.0 x 2) = 34.0

moles of H2O2 = $\frac{20}{34.0}$ = 0.588235…

1 L = 1 dm3

concentration = $\frac{0.588235...}{1}$ = 0.588235…

concentration = 0.59 mol dm-3 (2s.f.)

Practice Questions

1. Calculate the concentration, in mol dm-3, of the following solutions.

a. A solution formed by dissolving 0.5 moles of sodium chloride in 1 dm3 of water.

b. A solution formed by dissolving 0.125 moles of potassium manganate(VII) (KMnO4) in 100 cm3 of water.

c. A solution formed by dissolving 20.0 g of sodium hydroxide (NaOH) in 250 cm3 of water.

2. Calculate the number of moles of solute in each of the following solutions.

a. 1 dm3 of 0.5 mol dm-3 iron(II) sulfate (FeSO4) solution.

b. 100 cm3 of 1.5 mol dm-3 potassium chloride (KCl) solution.

c. 250 cm3 of 73 g dm-3 hydrochloric acid (HCl) solution

3. A student is preparing a 0.500 mol dm-3 solution of potassium nitrate (KNO3). Calculate the mass, in grams, of potassium nitrate the student should dissolve in
100 cm3 of water.

4. A student is preparing a 0.125 mol dm-3 solution of lithium hydroxide (LiOH). Calculate the volume of water, in cm3, which is needed to dissolve 2.39 g of lithium hydroxide to prepare this solution.

5. A student completely dissolved 49.05 g of an unidentified white solid in 500 cm3 of water. The student then carried out a titration and determined that a 25.0 cm3 sample of the solution contained 0.0250 moles of the solid. Calculate the relative molecular mass (Mr) of the solid.

1. Calculate the concentration, in mol dm-3, of the following solutions.

a. A solution formed by dissolving 0.5 moles of sodium chloride in 1 dm3 of water.

concentration = $\boldsymbol{\frac{0.5}{1}}$ =0.5 mol dm-3

b. A solution formed by dissolving 0.125 moles of potassium manganate(VII) (KMnO4) in 100 cm3 of water.

volume = $\boldsymbol{\frac{100\,\textbf{cm}^3}{1000}}$ = 0.1 dm3

concentration = $\boldsymbol{\frac{0.125}{0.1}}$ = 1.25 mol dm-3

c. A solution formed by dissolving 20.0 g of sodium hydroxide (NaOH) in 250 cm3 of water.

Mr of NaOH = (23.0 x 1) + (16.0 x 1) + (1.0 x 1) = 40

moles of NaOH = $\boldsymbol{\frac{20.0}{40.0}}$ = 0.5

volume = $\boldsymbol{\frac{250\,\textbf{cm}^3}{1000}}$ = 0.25 dm3

concentration = $\boldsymbol{\frac{0.5}{0.25}}$ = 2 mol dm-3

2. Calculate the number of moles of solute in each of the following solutions.

a. 1 dm3 of 0.5 mol dm-3 iron(II) sulfate (FeSO4) solution.

moles = 1 x 0.5 = 0.5

b. 100 cm3 of 1.5 mol dm-3 potassium chloride (KCl) solution.

volume = $\boldsymbol{\frac{100\,\textbf{cm}^3}{1000}}$ = 0.1 dm3

moles = 0.1 x 1.5 = 0.15

c. 250 cm3 of 73 g dm-3 hydrochloric acid (HCl) solution

Mr of HCl = (1.0 x 1) + (35.5 x 1) = 36.5

moles of HCl in 73g = $\boldsymbol{\frac{73}{36.5}}$ = 2

volume = $\boldsymbol{\frac{250\,\textbf{cm}^3}{1000}}$ = 0.25 dm3

moles = 0.25 x 2 = 0.5

3. A student is preparing a 0.500 mol dm-3 solution of potassium nitrate (KNO3). Calculate the mass, in grams, of potassium nitrate the student should dissolve in
100 cm3 of water.

volume of water = $\boldsymbol{\frac{100\,\textbf{cm}^3}{1000}}$ = 0.1 dm3

moles of KNO3 in prepared solution = 0.1 x 0.500 = 0.05

Mr of KNO3 = (39.1 x 1) + (14.0 x 1) + (16.0 x 3) + 101.1

mass = 0.05 x 101.1 = 5.055 = 5.06g (3s.f.)

4. A student is preparing a 0.125 mol dm-3 solution of lithium hydroxide (LiOH). Calculate the volume of water, in cm3, which is needed to dissolve 2.39 g of lithium hydroxide to prepare this solution.

Mr of LiOH = (6.9 x 1) + (16.0 x 1) + (1.0 x 1) = 23.9

moles of LiOH = $\boldsymbol{\frac{2.30}{23.9}}$ = 0.1 moles

volume = $\boldsymbol{\frac{\textbf{number of moles}}{\textbf{concentration}}=\frac{0.1}{0.125}}$ = 0.8dm3

0.8 dm3 x 1000 = 800 cm3

5. A student completely dissolved 49.05 g of an unidentified white solid in 500 cm3 of water. The student then carried out a titration and determined that a 25.0 cm3 sample of the solution contained 0.0250 moles of the solid. Calculate the relative molecular mass (Mr) of the solid.

volume of sample used in titration = $\boldsymbol{\frac{25.0 \, \textbf{cm}^3}{1000}}$ = 0.025 dm3

concentration = $\boldsymbol{\frac{0.0250}{0.025}}$ = 1 mol dm-3

volume of original solution = $\boldsymbol{\frac{500\,\textbf{cm}^3}{1000}}$ = 0.5 dm3

moles in original solution = 0.5 x 1 = 0.5

Mr = $\boldsymbol{\frac{\textbf{mass of substance}}{\textbf{number of moles}}=\frac{49.05}{0.5}}$ = 98.1

Found this A Level Chemistry blog post on Moles and Solutions useful? You can read our other Science revision blogs here.