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When a solute dissolves in a solvent it forms a solution.
The most common solvent is water. An aqueous solution forms when water is the solvent.
Solutes can be either a liquid or a solid. In the Chemistry lab at high school we usually use solid solutes weighed in grams (g).
A solution whose concentration is known precisely is called a standard solution.
There are four units of concentration that you are likely to come across in high school Chemistry.
Concentration in moles per litre, mol L-1
Concentration in grams per litre, g L-1
Percentage concentration, % concentration
Concentration in parts per million or milligrams per litre, mg L-1 or ppm
The concentration, c, of a solution is measured in moles per litre (mol L-1 ), and can be calculated using the following equation
c = n/V which can be re-arranged to n = cV and V = n/c
c = concentration in moles per litre, mol L-1
n = amount of dissolved solute in moles, mol.
V = volume of solution in litres, L
Some quantitative problems involving concentration may involve the calculation of the amount of substance, n (in mol) from mass, m (in g) and molar mass, M (in g mol-1 ) first.
This will mean that the n = m/M will need to be used, before the concentration (in mol L-1) can be calculated using c = n/V
Anhydrous (without water) sodium carbonate can be used to make a standard solution.
3.00 g of anhydrous sodium carbonate, Na2CO3 is dissolved in distilled water and the volume of the solution is made up to 250 mL in a volumetric flask. Calculate the concentration of the sodium carbonate solution in mol L-1.
M(Na2CO3) = (2 x 23.0) + 12.0 + (3 x 16.0) = 106 g mol-1
V(Na2CO3) = 250 mL = 0.250 L
Step 1: Calculate the number of moles of Na2CO3 using n = m/M.
n(Na2CO3) = m/M = 3.00/106 = 0.028301886 mol = 0.0283 mol (3 s.f.)
Step 2: Calculate the concentration of the solution using the unrounded number of moles from step 1.
c(Na2CO3) = n/V = 0.028301886/0.250 = 0.113207547 mol L-1
c(Na2CO3) = 0.113 mol L-1 (3 s.f.)
IMPORTANT: You will need to do one of these calculations in your assessment.
Concentration us also commonly expressed in terms of the mass of the solute in a litre of water.
The units for mass concentration are grams per litre, g L-1
To convert from mol L-1 to g L-1 just multiply the concentration in mol L-1 by the molar mass, M in g mol -1
Convert 0.113 mol L-1 sodium carbonate, Na2CO3 into g L-1
c(Na2CO3) = 0.113 mol L-1 & M(Na2CO3) = 106 g mol-1
mass concentration = 0.113 x 106 = 11.978 g L-1
Convert 0.123 g L-1 sodium carbonate, Na2CO3 , into mol L-1
mass concentration(Na2CO3) = 0.123 g L-1 & M(Na2CO3) = 106 g mol-1
c(Na2CO3) = 0.123/106 = 1.16 x 10-3 mol L-1 (3 s.f.)
IMPORTANT: You will need to do one of these calculations in your assessment.
The percentage concentration of a solution is often expressed as mass percent.
In Chemistry you might have to make a percentage concentration of a solution. This is quite easy to do. To make a 2 % solution, you would dissolve 2 g of the solute in 100 mL of distilled water. To make 200 mL of a 5 % solution, you would need to use 5 g per 100 mL of solution. Since 200 mL is needed you would need to have 2 x 5 g = 10 g of solute dissolved and made up to 200 mL with distilled water.
As a senior Chemistry student you might come across concentration in parts per million, or ppm.
This unit is often used when dealing with very small concentrations, such as oxygen concentration in water.
Parts per million, ppm, is the same as milligrams per litre, mg L-1
To calculate concentration in ppm, just multiply the mass concentration in g L-1 by 1000.
5.67 x 10-3 g L-1 = 5.67 x 10-3 x 1000 = 5.67 ppm
0.00789 g L-1 = 0.00789 x 1000 = 7.89 ppm
In the video below they refer to dm3 (this is the same as a litre, L), and cm3 (this is the same as a millilitre, mL)