Molar Concentrations of Liquids

Molar Concentrations of Liquids

A typical solution is made by dissolving some solid substance in a liquid. The amount of substance that is dissolved in a given volume of liquid is known as the concentration of the liquid. Mathematically, concentration (\(C\)) is defined as moles of solute (\(n\)) per unit volume (\(V\)) of solution.

\[C = \frac{n}{V}\]

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For this equation, the units for volume are \(\text{dm$^{3}$}\) (which is equal to litres). Therefore, the unit of concentration is \(\text{mol·dm$^{-3}$}\).

Definition: Concentration

Concentration is a measure of the amount of solute that is dissolved in a given volume of liquid. It is measured in \(\text{mol·dm$^{-3}$}\).

Example: Concentration Calculations

Question

If \(\text{3.5}\) \(\text{g}\) of sodium hydroxide (\(\text{NaOH}\)) is dissolved in \(\text{2.5}\) \(\text{dm$^{3}$}\) of water, what is the concentration of the solution in \(\text{mol·dm$^{-3}$}\)?

Step 1: Find the number of moles of sodium hydroxide

\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{3.5}\text{ g}}{\text{40.01}\text{ g·mol$^{-1}$}} \\ & = \text{0.0875}\text{ mol} \end{align*}

Step 2: Calculate the concentration

\begin{align*} C & = \frac{n}{V} \\ & = {\text{0.0875}\text{ mol}}{\text{2.5}\text{ dm$^{3}$}} \\ & = \text{0.035}\text{ mol·dm$^{-3}$} \end{align*}

The concentration of the solution is \(\text{0.035}\) \(\text{mol·dm$^{-3}$}\).

Example: Concentration Calculations

Question

You have a \(\text{1}\) \(\text{dm$^{3}$}\) container in which to prepare a solution of potassium permanganate (\(\text{KMnO}_{4}\)). What mass of \(\text{KMnO}_{4}\) is needed to make a solution with a concentration of \(\text{0.2}\) \(\text{mol·dm$^{-3}$}\)?

Step 1: Calculate the number of moles

\(C = \frac{n}{V}\) therefore:

\begin{align*} n & = \frac{C}{V} \\ & = \frac{\text{0.2}\text{ mol·dm$^{-3}$}}{\text{1}\text{ dm$^{3}$}} \\ & = \text{0.2}\text{ mol} \end{align*}

Step 2: Find the mass

\begin{align*} m & = nM \\ & = (\text{0.2}\text{ mol})(\text{158}\text{ g·mol$^{-1}$}) \\ & = \text{31.6}\text{ g} \end{align*}

The mass of \(\text{KMnO}_{4}\) that is needed is \(\text{31.6}\) \(\text{g}\).

Example: Concentration Calculations

Question

How much sodium chloride (in g) will one need to prepare \(\text{500}\) \(\text{cm$^{3}$}\) of solution with a concentration of \(\text{0.01}\) \(\text{mol·dm$^{-3}$}\)?

Step 1: Convert the given volume to the correct units

\[V = \text{500}\text{ cm$^{3}$} \times \frac{\text{1}\text{ dm$^{3}$}}{\text{1 000}\text{ cm$^{3}$}} = \text{0.5}\text{ dm$^{3}$}\]

Step 2: Find the number of moles

\begin{align*} n & = \frac{C}{V} \\ & = \frac{\text{0.01}\text{ mol·dm$^{-3}$}}{\text{0.5}\text{ dm$^{3}$}} \\ & = \text{0.005}\text{ mol} \end{align*}

Step 3: Find the mass

\begin{align*} m & = nM \\ & = (\text{0.005}\text{ mol})(\text{58.45}\text{ g·mol$^{-1}$}) \\ & = \text{0.29}\text{ g} \end{align*}

The mass of sodium chloride needed is \(\text{0.29}\) \(\text{g}\)

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