# 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.

$C = \frac{n}{V}$

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

This is a lesson from the tutorial, Quantitative Aspects of Chemical Change and you are encouraged to log in or register, so that you can track your progress.