Molar Concentration: Definition, Units, and Applications
- You’re making lemonade.
- You carefully measure sugar and water, adjusting the sweetness to your taste.
- If it’s too sweet, you add water; if it’s not sweet enough, you add more sugar.
What you’re doing is intuitively changing the concentration of sugar in the solution.
Molar Concentration: The Core Definition
Molar concentration
Molar concentration, commonly called molarity, measures how many moles of solute are dissolved in one cubic decimeter (dm³) of solution.
Mathematically, it is expressed as:
$$
C = \frac{n}{V}
$$ where:
- $C$ = molar concentration (in $\text{mol dm}^{-3}$)
- $n$ = amount of solute (in moles)
- $V$ = volume of the solution (in dm³)
This formula is fundamental for preparing solutions with precise concentrations and predicting reaction outcomes in solution-phase chemistry.
Why Use Molar Concentration?
- Molar concentration bridges the macroscopic (measurable quantities like volume) with the microscopic (the number of particles in a solution).
- For instance, knowing the concentration of a reactant helps calculate how much product a reaction will yield.
Always convert volume to dm³ (1 dm³ = 1,000 cm³) when using the molar concentration formula to avoid calculation errors.
Units of Concentration: Moles and Grams
Concentration can be expressed in various units depending on the context. Let’s explore the two most common: $\text{mol dm}^{-3}$ and $\text{g dm}^{-3}$.
Molar Concentration $C$ in $\text{mol dm}^{-3}$:
This unit specifies the number of moles of solute per cubic decimeter of solution.
Dissolving 1 mole of sodium chloride $NaCl$ in 1 dm³ of water results in a concentration of $1.0 \text{mol dm}^{-3}$.
Mass Concentration $\rho$ in $\text{g dm}^{-3}$:
Mass concentration
Mass concentration specifies the mass of solute (in grams) dissolved in one cubic decimeter (dm³) of solution.
- It focuses on the actual mass rather than the number of particles, making it useful in laboratory preparations when measuring mass directly is easier than calculating moles: $$\rho = \frac{m}{V}$$ where:
- $\rho$ = mass concentration ($\text{g dm}^{-3}$)
- $m$ = mass of solute (g)
- $V$ = volume of solution (dm³)
- Since the number of moles $n$ is related to mass by the molar mass $M$ ($n = \frac{m}{M}$), the mass concentration can also be calculated from molar concentration: $$\rho = C \times M$$ where:
- $C$ = molar concentration ($\text{mol dm}^{-3}$)
- $M$ = molar mass of the solute ($\text{g mol}^{-1}$)
Converting Between Units
- A solution of 0.5 $mol \, dm^{-3}$ sodium chloride $NaCl$ has a molar mass of 58.44 $mol \, dm^{-3}$.
- To find its mass concentration: $$\rho = C \times M = 0.5 \, \text{mol dm}^{-3} \times 58.44 \, \text{g mol}^{-1} = 29.22 \, \text{g dm}^{-3} $$
- This solution contains $29.22 \, \text{g}$ of $NaCl$ per dm³.
- A common mistake is forgetting to convert volume to dm³ when using $C = \frac{n}{V}$.
- For instance, if the volume is given in cm³, divide it by 1,000 to convert to dm³.
Molarity
Applications of Molar Concentration
- Molar concentration is a practical tool used in laboratory experiments, industrial processes, and even daily activities.
- Below are examples of common problem types involving molar concentration.
Calculating molar concentration
You dissolve $2.00 \, \text{mol}$ of glucose $C_6H_{12}O_6$ in $5.00 \, \text{dm}^3$ of water. What is the molar concentration of glucose in the solution?
Solution
Using $C = \frac{n}{V}$:
$$
C = \frac{2.00 \, \text{mol}}{5.00 \, \text{dm}^3} = 0.400 \, \text{mol dm}^{-3}
$$
The concentration of glucose is $0.400 \, \text{mol dm}^{-3}$.
Finding the Amount of Solute
How many moles of sodium hydroxide $NaOH$ are present in $250 \, \text{cm}^3$ of a $0.200 \, \text{mol dm}^{-3}$ solution?
Solution
Convert the volume to dm³:
$$
V = \frac{250 \, \text{cm}^3}{1,000} = 0.250 \, \text{dm}^3
$$
Now use $n = C \times V$:
$$
n = 0.200 \, \text{mol dm}^{-3} \times 0.250 \text{dm}^3 = 0.0500 \, \text{mol}
$$
The solution contains $0.0500 \, \text{mol}$ of $NaOH$.
Determining Solution Volume
What volume of a $2.00 \, \text{mol dm}^{-3}$ sulfuric acid $H_2SO_4$ solution is required to provide $1.00 \, \text{mol}$ of solute?
Solution
Rearrange $C = \frac{n}{V}$ to solve for $V$:
$$
V = \frac{n}{C} = \frac{1.00 \, \text{mol}}{2.00 \, \text{mol dm}^{-3}} = 0.500 \, \text{dm}^3
$$
You need $0.500 \, \text{dm}^3$ or $500 \, \text{cm}^3$ of the solution.
A solution has a concentration of $0.100 \, \text{mol dm}^{-3}$. If you have $50.0 \, \text{cm}^3$ of it, how many moles of solute does it contain?
Practical Implications
Dilutions
- Molar concentration is essential when preparing solutions of specific concentrations.
- To dilute a stock solution, use the formula:$$C_1 V_1 = C_2 V_2
$$ where:- $C_1$ and $V_1$: concentration and volume of the stock solution
- $C_2$ and $V_2$: concentration and volume of the diluted solution
You have $100 \, \text{cm}^3$ of a $2.00 \, \text{mol dm}^{-3}$ hydrochloric acid $HCl$ solution. How much water must you add to dilute it to $0.500 \, \text{mol dm}^{-3}$?
Solution
- Using $C_1 V_1 = C_2 V_2$:
$$
2.00 \, \text{mol dm}^{-3})(100 \, \text{cm}^3) = (0.500 \, \text{mol dm}^{-3})(V_2)
$$
$$
V_2 = \frac{200.0 \, \text{mol cm}^3}{0.500 \, \text{mol dm}^{-3}} = 400 \, \text{cm}^3
$$
- The final solution volume is $400 \, \text{cm}^3$.
- Add $400 - 100 = 300 \, \text{cm}^3$ of water.
