Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 1 - 1.5 Concentration-Study Notes - New Syllabus

1.5 Concentration of Solutions

The concentration of a solution describes how much solute is dissolved in a given volume of solution. In chemistry, concentration is commonly expressed in two ways: mol dm\(^{-3}\) and g dm\(^{-3}\).

At this stage, calculations involving titration are not required. Only basic concentration calculations are needed.

Concentration in \( \mathrm{mol\ dm^{-3}} \)

Concentration in \( \mathrm{mol\ dm^{-3}} \) (moles per cubic decimetre) is the number of moles of solute dissolved in one cubic decimetre of solution.

The concentration of a solution can be calculated using the relationship:

\( c = \dfrac{n}{V} \)

where:

• \( c \) = concentration \( (\mathrm{mol\ dm^{-3}}) \)
• \( n \) = number of moles \( (\mathrm{mol}) \)
• \( V \) = volume of solution \( (\mathrm{dm^3}) \)

Example

A solution contains \( 0.50 \) mol of sodium chloride dissolved in \( 2.0\ \mathrm{dm^3} \) of solution. Calculate the concentration.

\( c = \dfrac{0.50}{2.0} \)

\( c = 0.25\ \mathrm{mol\ dm^{-3}} \)

Concentration in \( \mathrm{g\ dm^{-3}} \)

Concentration in \( \mathrm{g\ dm^{-3}} \) is the mass of solute dissolved in one cubic decimetre of solution.

The relationship used is:

\( \text{concentration} = \dfrac{\text{mass of solute}}{\text{volume of solution}} \)

\( c = \dfrac{m}{V} \)

where:

• \( m \) = mass of solute (g)
• \( V \) = volume of solution (dm\(^{3}\))

Example

A solution contains \( 10\ \mathrm{g} \) of sodium chloride dissolved in \( 2.0\ \mathrm{dm^3} \) of solution. Calculate the concentration in \( \mathrm{g\ dm^{-3}} \).

\( c = \dfrac{10}{2.0} \)

\( c = 5.0\ \mathrm{g\ dm^{-3}} \)

Relationship Between the Two Units

Concentration in \( \mathrm{g\ dm^{-3}} \) can be converted to \( \mathrm{mol\ dm^{-3}} \) using the molar mass.

\( c(\mathrm{mol\ dm^{-3}}) = \dfrac{c(\mathrm{g\ dm^{-3}})}{\text{molar mass}} \)