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

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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:

For example:

1. Calculate the concentration of a solution that contains 0.5 moles of potassium iodide dissolved in 1.0 dm^{3} of water.

concentration = = 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 = = 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:

For example:

1. Calculate the number of moles of copper chloride in 1.0 dm^{3} 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 dm^{3} 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 (cm^{3}) rather than cubic decimetres (dm^{3}) 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.

Example Question 1

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

M_{r} of KCl = (39.1 x 1) + (35.5 x 1) = 74.6

moles of KCl = = 0.5

volume = = 0.5dm^{3}

concentration = = 1 mol dm^{-3}

Example Question 2

**Calculate the mass, in grams, of iron(II) sulfate (FeSO _{4}) in 50 cm^{3} of 0.25 mol dm^{-3} iron(II) sulfate solution.**

volume = =0.05 dm^{3}

number of moles of FeSO_{4} in solution = 0.05 x 0.25 = 0.0125

Mr of FeSO_{4} = (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 (H_{2}O_{2}) in mol dm^{-3}.**

M_{r} of H_{2}O_{2} = (1.0 x 2) + (16.0 x 2) = 34.0

moles of H_{2}O_{2} = = 0.588235…

1 L = 1 dm^{3}

concentration = = 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 dm^{3} of water.

b. A solution formed by dissolving 0.125 moles of potassium manganate(VII) (KMnO_{4}) in 100 cm^{3} of water.

c. A solution formed by dissolving 20.0 g of sodium hydroxide (NaOH) in 250 cm^{3} of water.

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

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

b. 100 cm^{3} of 1.5 mol dm^{-3} potassium chloride (KCl) solution.

c. 250 cm^{3 }of 73 g dm^{-3} hydrochloric acid (HCl) solution

3. A student is preparing a 0.500 mol dm^{-3} solution of potassium nitrate (KNO_{3}). Calculate the mass, in grams, of potassium nitrate the student should dissolve in

100 cm^{3} of water.

4. A student is preparing a 0.125 mol dm^{-3} solution of lithium hydroxide (LiOH). Calculate the volume of water, in cm^{3}, 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 cm^{3} of water. The student then carried out a titration and determined that a 25.0 cm^{3} sample of the solution contained 0.0250 moles of the solid. Calculate the relative molecular mass (M_{r}) of the solid.

Answers

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