Colligative properties

Study notes for CBSE Class 12 chemistry - Colligative properties.

• Colligative properties depend only on the number of solute particles, not their nature
• Relative lowering of vapour pressure: $\frac{\Delta P}{P^0} = x_2$ (mole fraction of solute)
• Boiling point elevation: $\Delta T_b = i \cdot K_b \cdot m$
• Freezing point depression: $\Delta T_f = i \cdot K_f \cdot m$
• Osmotic pressure: $\pi = i \cdot CRT$
• Van’t Hoff factor: $i = \frac{\text{observed colligative property}}{\text{calculated colligative property}}$

Problem: $34.2 \, \text{g}$ of sucrose ($M = 342 \, \text{g/mol}$) is dissolved in $500 \, \text{g}$ of water. Calculate the boiling point of the solution. ($K_b$ for water $= 0.52 \, \text{K}\cdot\text{kg/mol}$)

Solution:

Moles of sucrose:

$$n = \frac{34.2}{342} = 0.1 \, \text{mol}$$

Molality:

$$m = \frac{0.1}{0.5} = 0.2 \, \text{mol/kg}$$

Sucrose is a non-electrolyte, so $i = 1$.

Boiling point elevation:

$$\Delta T_b = i \cdot K_b \cdot m = 1 \times 0.52 \times 0.2 = 0.104 \, \text{K}$$

New boiling point:

$$T_b = 100 + 0.104 = 100.104^\circ\text{C}$$

Problem: Calculate the freezing point of a solution containing $5.85 \, \text{g}$ of NaCl ($M = 58.5 \, \text{g/mol}$) dissolved in $1 \, \text{kg}$ of water. ($K_f$ for water $= 1.86 \, \text{K}\cdot\text{kg/mol}$)

Solution:

Moles of NaCl:

$$n = \frac{5.85}{58.5} = 0.1 \, \text{mol}$$

Molality:

$$m = \frac{0.1}{1} = 0.1 \, \text{mol/kg}$$

NaCl dissociates completely: $\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$, so $i = 2$.

Freezing point depression:

$$\Delta T_f = i \cdot K_f \cdot m = 2 \times 1.86 \times 0.1 = 0.372 \, \text{K}$$

New freezing point:

$$T_f = 0 - 0.372 = -0.372^\circ\text{C}$$

Problem: $1.5 \, \text{g}$ of a polymer ($M = 150{,}000 \, \text{g/mol}$) is dissolved in $500 \, \text{mL}$ of water at $27^\circ\text{C}$. Find the osmotic pressure. ($R = 0.0821 \, \text{L}\cdot\text{atm/K}\cdot\text{mol}$)

Solution:

Moles of polymer:

$$n = \frac{1.5}{150{,}000} = 1 \times 10^{-5} \, \text{mol}$$

Concentration:

$$C = \frac{n}{V} = \frac{1 \times 10^{-5}}{0.5} = 2 \times 10^{-5} \, \text{mol/L}$$

Temperature: $T = 27 + 273 = 300 \, \text{K}$

Osmotic pressure:

$$\pi = CRT = 2 \times 10^{-5} \times 0.0821 \times 300$$ $$= 4.926 \times 10^{-4} \, \text{atm} \approx 0.000493 \, \text{atm}$$

Forgetting the van’t Hoff factor for electrolytes. NaCl dissociates into 2 ions (i = 2), CaCl₂ into 3 ions (i = 3). Using i = 1 for electrolytes halves or thirds the correct colligative property value. Always check whether the solute dissociates before applying formulas.

Confusing molality with molarity in colligative property formulas. Colligative properties use molality (moles per kg solvent), not molarity (moles per liter solution). Molarity changes with temperature while molality doesn’t, which is why colligative properties prefer molality. Students often grab the wrong concentration unit.

Using R = 0.0821 with SI pressure units. If pressure is in Pa or kPa, use R = 8.314 J/(mol·K). If pressure is in atm, use R = 0.0821 L·atm/(mol·K). Mixing these gives nonsensical answers. Always match R to your pressure units.

1. $9 \, \text{g}$ of urea ($M = 60 \, \text{g/mol}$) is dissolved in $500 \, \text{g}$ of water. Find the boiling point elevation. ($K_b = 0.52 \, \text{K}\cdot\text{kg/mol}$)
2. $11.7 \, \text{g}$ of NaCl is dissolved in $1 \, \text{kg}$ of water. Find the freezing point of the solution. ($K_f = 1.86 \, \text{K}\cdot\text{kg/mol}$)
3. A solution has an osmotic pressure of $2.46 \, \text{atm}$ at $300 \, \text{K}$. Find the molarity.

1. $1 \, \text{g}$ of $\text{CaCl}_2$ ($M = 111 \, \text{g/mol}$) is dissolved in $100 \, \text{g}$ of water. Calculate the freezing point depression. ($K_f = 1.86 \, \text{K}\cdot\text{kg/mol}$, assume complete dissociation).
2. A solution of glucose ($M = 180 \, \text{g/mol}$) has a freezing point of $-0.93^\circ\text{C}$. Find the molality and the mass of glucose in $500 \, \text{g}$ of water. ($K_f = 1.86 \, \text{K}\cdot\text{kg/mol}$)

Colligative properties depend only on how many solute particles you add, not what they are. Adding salt to water raises the boiling point and lowers the freezing point because solute particles get in the way of molecules trying to escape into the gas phase or lock into a crystal. Osmotic pressure is nature’s way of balancing concentrations across a membrane — water flows toward the more concentrated side until pressures equalize. The van’t Hoff factor accounts for the fact that electrolytes like NaCl split into multiple ions, so they have roughly double the effect of a sugar molecule that stays intact.

• Solutions: Colligative properties are direct applications of solution concentration concepts like molality and mole fraction.
• Chemical Kinetics: Solute concentration affects reaction rates, understanding colligative properties deepens your grasp of concentration effects.
• Biomolecules: Osmotic pressure determines how water moves across cell membranes, directly connecting colligative properties to biology.
• Surface Chemistry: Adsorption and colligative properties both arise from solute-solvent interactions at the molecular level.

• Solutions