IB MYP 4-5 Chemistry -Concentration of solutions (mol/dm³)- Study Notes - New Syllabus
IB MYP 4-5 Chemistry -Concentration of solutions (mol/dm³)- Study Notes
Key Concepts
- Concentration of Solutions (mol/dm³), Molarity, and Molality
Concentration of Solutions (mol/dm³), Molarity, and Molality
Concentration of Solutions (mol/dm³), Molarity, and Molality
The concentration of a solution expresses how much solute is dissolved in a certain amount of solvent or solution. It provides a quantitative measure of the “strength” or “dilution” of a solution.
Key Idea: Concentration can be expressed in several ways — most commonly as molarity (mol/dm³) and molality (mol/kg).
Concentration in mol/dm³ (Molar Concentration)
Formula:
\( \mathrm{c = \dfrac{n}{V}} \)
- \( \mathrm{c} \) = concentration (mol/dm³)
- \( \mathrm{n} \) = number of moles of solute (mol)
- \( \mathrm{V} \) = volume of solution (dm³)
Rearranged Forms:
- \( \mathrm{n = c \times V} \)
- \( \mathrm{V = \dfrac{n}{c}} \)
Important Conversion:
\( \mathrm{1\ dm^3 = 1000\ cm^3 = 1\ L} \)
So if the volume is in cm³, convert it using:
\( \mathrm{V(dm^3) = \dfrac{V(cm^3)}{1000}} \)
Molarity (M)
Molarity is the number of moles of solute dissolved in one cubic decimeter (1 dm³) of solution.
\( \mathrm{M = \dfrac{n}{V}} \)
- \( \mathrm{M} \) = molarity (mol/L or mol/dm³)
- \( \mathrm{n} \) = moles of solute
- \( \mathrm{V} \) = volume of solution (L or dm³)
Unit: mol/L or mol/dm³
Note: Molarity changes with temperature because solution volume expands or contracts.
Molality (m)
Molality is the number of moles of solute dissolved in 1 kilogram of solvent.
\( \mathrm{m = \dfrac{n}{m_{solvent}(kg)}} \)
- \( \mathrm{m} \) = molality (mol/kg)
- \( \mathrm{n} \) = number of moles of solute
- \( \mathrm{m_{solvent}} \) = mass of solvent (kg)
Unit: mol/kg
Note: Molality is independent of temperature because it depends on mass, not volume.
Relationship Between Molarity, Molality, and Mass
Since \( \mathrm{n = \dfrac{m}{M}} \), we can express molarity and molality as:
\( \mathrm{M = \dfrac{m}{M_r \times V}} \)
\( \mathrm{m = \dfrac{m}{M_r \times mass_{solvent}(kg)}} \)
- \( \mathrm{m} \) = mass of solute (g)
- \( \mathrm{M_r} \) = molar mass (g/mol)
- \( \mathrm{V} \) = volume of solution (dm³)
Comparison Between Molarity and Molality
| Property | Molarity (M) | Molality (m) |
|---|---|---|
| Definition | Moles of solute per dm³ of solution | Moles of solute per kg of solvent |
| Formula | \( \mathrm{M = \dfrac{n}{V}} \) | \( \mathrm{m = \dfrac{n}{mass_{solvent}(kg)}} \) |
| Depends on | Volume of solution | Mass of solvent |
| Affected by temperature? | Yes (volume changes) | No (mass constant) |
| Commonly used in | Chemical reactions, titrations | Colligative property calculations |
Summary of Key Formulas
| Quantity | Formula | Unit |
|---|---|---|
| Concentration (molarity) | \( \mathrm{c = \dfrac{n}{V}} \) | mol/dm³ |
| Molality | \( \mathrm{m = \dfrac{n}{mass_{solvent}(kg)}} \) | mol/kg |
| Moles | \( \mathrm{n = c \times V} \) | mol |
Example
Find the molarity of a solution containing 0.5 mol of sodium chloride in 0.25 dm³ of solution.
▶️ Answer / Explanation
Step 1: \( \mathrm{M = \dfrac{n}{V}} \)
Step 2: \( \mathrm{M = \dfrac{0.5}{0.25} = 2.0\ mol/dm^3} \)
Final Answer: \( \mathrm{2.0\ mol/dm^3} \)
Example
Calculate the molality of a solution made by dissolving 18 g of glucose (\( \mathrm{C_6H_{12}O_6} \)) in 90 g of water. \( \mathrm{M_r(C_6H_{12}O_6) = 180} \)
▶️ Answer / Explanation
Step 1: \( \mathrm{n = \dfrac{m}{M_r} = \dfrac{18}{180} = 0.1\ mol} \)
Step 2: \( \mathrm{mass_{solvent} = 90\ g = 0.09\ kg} \)
Step 3: \( \mathrm{m = \dfrac{0.1}{0.09} = 1.11\ mol/kg} \)
Final Answer: \( \mathrm{1.11\ molal} \)
Example
A solution is made by dissolving 9.8 g of sulfuric acid (\( \mathrm{H_2SO_4} \)) in 200 cm³ of water (density = 1 g/cm³). Calculate both the molarity and molality of the solution. \( \mathrm{M_r(H_2SO_4) = 98} \)
▶️ Answer / Explanation
Step 1: Moles of solute = \( \mathrm{n = \dfrac{9.8}{98} = 0.1\ mol} \)
Step 2: Volume of solution = \( \mathrm{200\ cm^3 = 0.2\ dm^3} \)
Step 3: Molarity \( \mathrm{M = \dfrac{n}{V} = \dfrac{0.1}{0.2} = 0.5\ mol/dm^3} \)
Step 4: Mass of solvent ≈ \( \mathrm{200\ g = 0.2\ kg} \)
Step 5: Molality \( \mathrm{m = \dfrac{0.1}{0.2} = 0.5\ mol/kg} \)
Final Answer: Molarity = \( \mathrm{0.5\ mol/dm^3} \), Molality = \( \mathrm{0.5\ mol/kg} \)