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CSEC Chemistry: The Mole in Relation to Solutions

You may already be familiar with the description of a solution as dilute, concentrated or saturated. However, with the mole, we have a far more precise way to state the concentration of a solution based on how much solute is dissolved.

We will discuss two ways of expression concentration:

  • Mass concentration, i.e. the mass of solute dissolved in 1000 cm^3 of solution. This is expressed in g/dm^3.

  • Molar concentration, or molarity, i.e. the number of moles in 1000 cm^3 of solution. This is expressed in mol/dm^3.

Mass Concentration

Example 1: Find the mass concentration of 200g of potassium hydroxide in 5 dm^3 of solution.

We simply divide the mass of solute by the number of dm^3 of solution to find the mass of solute in 1 dm^3 of solution:

1 dm^3 of KOH (aq) contains (200/5)g= 40 g/dm^3

Example 2: Find the mass concentration of 10 dm^3 of a solution containing 1 kg copper (II) sulphate.

1 dm^3 of CuSO4(aq) contains (1000g/10)g= 100 g/dm^3

Molar Concentration

Example 1: Find the molar concentration of a solution containing 0.015 mol of HCl in 25 cm^3 of solution.

First, you convert the cm^3 volume to dm^3 by dividing by 1000.

25 cm^3/1000= 0.025 dm^3

You then divide the number of moles by the dm^3 value to give you the number of moles in 1 dm^3 of solution.

1 dm^3 of HCl(aq) contains (0.015 mol/0.025 dm^3)= 0.6 mol/dm^3

Example 2: Find the molar concentration of 10.0 mol of NaOH in 7.2 dm^3 of solution.

1 dm^3 of NaOH(aq) contains (10.0 mol/7.2 dm^3)= 1.39 mol/dm^3

If you have the molar concentration of a solution, you can find the number of moles present in any volume of said solution.

number of moles in volume, V(cm^3)= (V/1000) dm^3 x concentration in mol/dm^3

Example: Calculate the number of moles of solute in 25 cm^3 of 0.05 mol/dm^3 H2SO4.

number of moles= (25 cm^3/1000) x 0.05 mol/dm^3= 0.00125 mol

It is also possible to convert between mass concentration and molar concentration.

molar concentration = mass concentration of solute (g/dm^3)/molar mass of solute

Transposition of this expression gives you two other relationships:

mass concentration of solute= molar concentration x molar mass of solute

molar mass of solute=mass concentration of solute/molar concentration of solute

Example: What is the molar concentration or molarity (mol/dm^3) of a solution of nitric acid containing 2.52 g/dm^3 HNO3?

The relative molecular mass of nitric acid= [1 + 14 + (3 x 16)]= 63

(We covered relative molecular mass and how it is numerically equal to the molar mass in this lesson)

molarity= mass concentration of nitric acid/molar mass of nitric acid

=2.52 gdm^-3/63 gmol^-1= 0.04 mol/dm^3

Example 2: 250 cm^3 of a solution of sodium chloride contains 11.70 g NaCl. Find the concentration in mol/dm^3 of this solution.

We first find the mass concentration in g/dm^3:

0.25 dm^3 of solution contains 11.70 of solute

1 dm^3 of solution contains (11.70/0.25)g= 46.8 g/dm^3

We then convert this to molarity:

(remember that the molar mass of NaCl= 23+35.5= 58.5 g/mol

molarity= mass concentration/molar mass

= 46.8 gdm^-3/58.5 gmol^-1 = 0.80 mol/dm^3

Example 3: Given a solution of MCl2 has a molarity of 0.025 moldm^-3 and a mass concentration of 2.375 gdm^-3, find the relative atomic mass of M?

You first find the molar mass:

molar mass of MCl2= mass concentration/molarity

= 2.375 gdm^-3/0.025 moldm^-3

= 95.0 g/mol

You then calculate the relative atomic mass of M:

relative atomic mass of M + relative atomic mass of Cl= relative atomic mass of MCl2

M + [2 x 35.5] = 95

M = 95- 71

relative atomic mass of M= 24

You could go on to deduce that the element M is Magnesium.

Volumetric Analysis

This is a method of determining the concentrations of solutions. The main aspect of volumetric analysis is measuring the volume of one solution with an accurately known concentration which is required to react quantitatively with a solution of the substance being determined.

The solution with an accurately known concentration in the volumetric analysis is called the standard solution.

The apparatus commonly used in volumetric analysis includes a balance, a pipette, a burette and a volumetric flask.

The standard solution is usually added to a titration flask via a pipette. The solution whose concentration is to be determined in the analysis (also known as the analyte) is then added to the standard solution from a burette. The process of adding solution until the reaction is complete is known as a titration.

The point where the reaction is complete is called the equivalence point or end-point, and the volume of solution added from the burette is the titre.

(You can find a more detailed explanation of the procedure of volumetric analysis in this Wired Chemist article)

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Hello Chemistry students! This material was covered in Quelpr's CSEC Chemistry on June 26, 2021, however the past paper questions and solutions have been removed. If you are interested in similar cont

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