Study Notes on Molality and Solution Concentration

Solutions and Concentration Units

Molality

Molality is a concentration unit used in chemistry to express the amount of solute dissolved in a given mass of solvent. It is particularly useful when studying properties that depend on the number of particles in solution, such as colligative properties.

• Definition: Molality (m) is defined as the number of moles of solute per kilogram of solvent.

• Formula:

$\text{Molality} = \frac{\text{Moles of solute}}{\text{kg of solvent}}$

• Example Calculation: If 0.30 moles of NaCl are dissolved in 1 kg of water, the molality is:

$0.30\ \text{mol NaCl} / 1\ \text{kg H}_2\text{O} = 0.30\ \text{molal}$

Worked Example: Calculating Molality

Suppose a solution is prepared by dissolving 43.0 g of potassium nitrate (KNO3) in enough water to make 100.0 mL of solution. The density of the solution is 1.19 g/mL. What is the molality of KNO3 in the solution?

• Step 1: Calculate moles of KNO3:

  • Molar mass of KNO3 = 101.1 g/mol

  • Moles = $\frac{43.0\ \text{g}}{101.1\ \text{g/mol}} = 0.426\ \text{mol}$

• Step 2: Calculate mass of solvent (water):

  • Total mass of solution = $100.0\ \text{mL} \times 1.19\ \text{g/mL} = 119\ \text{g}$

  • Mass of solvent = Total mass - Mass of solute = $119\ \text{g} - 43.0\ \text{g} = 76.0\ \text{g} = 0.0760\ \text{kg}$

• Step 3: Calculate molality:

  • $\text{Molality} = \frac{0.426\ \text{mol}}{0.0760\ \text{kg}} = 5.61\ \text{molal}$

Additional Example: Molality from Density

Find the molality of a solution of glucose (C6H12O6) in an aqueous solution that is 2.5 M. The density of the solution is 1.15 g/mL.

• Step 1: Calculate moles of glucose in 1 L of solution:

  • Moles = $2.5\ \text{mol}$

• Step 2: Calculate mass of solution:

  • Mass = $1000\ \text{mL} \times 1.15\ \text{g/mL} = 1150\ \text{g}$

• Step 3: Calculate mass of solvent:

  • Mass of glucose = $2.5\ \text{mol} \times 180.2\ \text{g/mol} = 450.5\ \text{g}$

  • Mass of solvent = $1150\ \text{g} - 450.5\ \text{g} = 699.5\ \text{g} = 0.6995\ \text{kg}$

• Step 4: Calculate molality:

  • $\text{Molality} = \frac{2.5\ \text{mol}}{0.6995\ \text{kg}} = 3.57\ \text{molal}$

Practice Problem

What is the mass percent of Na2 in a 1.25 m aqueous solution of Na2?

• Answer: 3.51%

Summary Table: Molality vs. Molarity

Concentration Unit	Definition	Formula	Depends on
Molality (m)	Moles of solute per kg of solvent	$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$	Mass of solvent
Molarity (M)	Moles of solute per liter of solution	$M = \frac{\text{moles of solute}}{\text{L of solution}}$	Volume of solution

Additional info: Molality is preferred over molarity when temperature changes are involved, as mass does not vary with temperature, while volume can expand or contract.