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.

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

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 $\text{0.29}$ $\text{g}$

Molar Concentrations of Liquids