Depression of Freezing Point — Revision Notes

⚡ 30-Second Revision

Definition: — Decrease in freezing point of solvent upon adding non-volatile solute.
Formula: — $\Delta T_f = i \cdot K_f \cdot m$
$\Delta T_f$ — Depression in freezing point ($T_f^0 - T_f^{\text{solution}}$).
$i$ — Van't Hoff factor (number of particles produced per formula unit).

- Non-electrolytes: $i=1$ (e.g., glucose, urea). - Electrolytes: $i > 1$ (e.g., NaCl $i=2$, CaCl$_2$ $i=3$).

$K_f$ — Cryoscopic constant (molal freezing point depression constant), solvent-specific. For water, $K_f = 1.86,\text{K kg/mol}$.
$m$ — Molality (moles of solute per kg of solvent). $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$.
Applications: — Antifreeze, de-icing, molar mass determination (cryoscopy).

2-Minute Revision

Depression of freezing point is a colligative property, meaning it depends on the number of solute particles, not their identity. When a non-volatile solute is added to a solvent, the freezing point of the solution becomes lower than that of the pure solvent.

This occurs because the solute particles interfere with the solvent molecules' ability to form an ordered solid structure, requiring a lower temperature. Quantitatively, the depression ($\Delta T_f$) is directly proportional to the molality ($m$) of the solution: $\Delta T_f = i \cdot K_f \cdot m$.

Here, $K_f$ is the cryoscopic constant, unique to each solvent, and $i$ is the Van't Hoff factor, accounting for the number of particles produced by the solute (e.g., $i=1$ for non-electrolytes, $i=2$ for NaCl).

To solve problems, first calculate $i$, then the molality ($m = \text{moles of solute} / \text{kg of solvent}$), and finally apply the formula. Remember to subtract $\Delta T_f$ from the pure solvent's freezing point to get the solution's freezing point.

This concept is vital for determining unknown molar masses and understanding real-world applications like antifreeze.

5-Minute Revision

Depression of freezing point is a crucial colligative property, defined as the lowering of a solvent's freezing point when a non-volatile solute is dissolved in it. The underlying reason is the reduction in the solvent's vapor pressure by the solute, making the liquid phase more stable at lower temperatures relative to the solid phase. The pure solvent solidifies out, leaving a more concentrated solution. The mathematical relationship is given by $\Delta T_f = i \cdot K_f \cdot m$.

Let's break down the terms:

$\Delta T_f$ — The observed depression in freezing point, calculated as $T_f^0 - T_f^{\text{solution}}$, where $T_f^0$ is the freezing point of the pure solvent (e.g., $0^circ C$ for water) and $T_f^{\text{solution}}$ is the freezing point of the solution.
$i$ (Van't Hoff factor) — This accounts for the effective number of particles in solution. For non-electrolytes (like glucose, urea), $i=1$. For electrolytes that dissociate, $i$ equals the number of ions formed (e.g., NaCl $\rightarrow$ Na$^+$ + Cl$^-$, so $i=2$; K$_2$SO$_4 \rightarrow$ 2K$^+$ + SO$_4^{2-}$, so $i=3$). If association occurs, $i < 1$.
$K_f$ (Cryoscopic Constant) — This is a characteristic constant for a specific solvent, representing the freezing point depression for a 1 molal solution. For water, $K_f = 1.86,\text{K kg/mol}$. It's independent of the solute.
$m$ (Molality) — Concentration expressed as moles of solute per kilogram of solvent. It's preferred over molarity because it's temperature-independent: $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$.

Example: Calculate the freezing point of a solution containing $12,\text{g}$ of urea (molar mass $60,\text{g/mol}$) in $250,\text{g}$ of water. ($K_f$ for water = $1.86,\text{K kg/mol}$).

1
Identify $i$ — Urea is a non-electrolyte, so $i=1$.
2
Calculate moles of urea — $n_{\text{urea}} = 12,\text{g} / 60,\text{g/mol} = 0.2,\text{mol}$.
3
Calculate molality — $W_{\text{water}} = 250,\text{g} = 0.250,\text{kg}$. $m = 0.2,\text{mol} / 0.250,\text{kg} = 0.8,\text{mol/kg}$.
4
Calculate $\Delta T_f$ — $\Delta T_f = i \cdot K_f \cdot m = 1 \cdot 1.86,\text{K kg/mol} \cdot 0.8,\text{mol/kg} = 1.488,\text{K}$.
5
Calculate solution freezing point — $T_f^{\text{solution}} = T_f^0 - \Delta T_f = 0^circ C - 1.488^circ C = -1.488^circ C$.

This property is widely used in antifreeze solutions (e.g., ethylene glycol in car radiators) and for de-icing roads with salt. It's also a powerful analytical tool (cryoscopy) for determining the molar mass of unknown non-volatile solutes.

Prelims Revision Notes

Depression of Freezing Point ($\Delta T_f$)

1. Definition & Concept:

Freezing Point: — Temperature at which liquid and solid phases are in equilibrium.
Depression: — Lowering of freezing point of a solvent when a non-volatile solute is added.
Reason: — Solute particles interfere with solvent's crystallization process, requiring lower temperature for solidification. Also, solute lowers vapor pressure of liquid solvent, making it equal to pure solid solvent's vapor pressure at a lower temperature.
Colligative Property: — Depends only on the *number* of solute particles, not their identity.

2. Key Formula:

$\Delta T_f = i \cdot K_f \cdot m$

3. Terms Explained:

$\Delta T_f$ (Depression in Freezing Point):**

* Calculated as $T_f^0 - T_f^{\text{solution}}$. * $T_f^0$: Freezing point of pure solvent (e.g., $0^circ C$ or $273.15,\text{K}$ for water). * $T_f^{\text{solution}}$: Freezing point of the solution. * Units: K or $^circ C$.

$i$ (Van't Hoff Factor):**

* Accounts for dissociation/association of solute. * Non-electrolytes: $i=1$ (e.g., glucose, urea, sucrose). * Electrolytes (complete dissociation): $i = \text{number of ions per formula unit}$.

* NaCl: $i=2$ (Na$^+$ + Cl$^-$) * CaCl$_2$: $i=3$ (Ca$^{2+}$ + 2Cl$^-$) * K$_3$[Fe(CN)$_6$]: $i=4$ (3K$^+$ + [Fe(CN)$_6$]$^{3-}$) * Incomplete dissociation/association: $i$ will be between 1 and the theoretical maximum (for dissociation) or less than 1 (for association).

Often calculated from experimental data.

$K_f$ (Cryoscopic Constant / Molal Freezing Point Depression Constant):**

* A constant characteristic of the *solvent*. * Represents $\Delta T_f$ for a 1 molal solution. * Units: K kg/mol or $^circ C$ kg/mol. * For water: $K_f = 1.86,\text{K kg/mol}$.

$m$ (Molality):**

* Concentration unit: moles of solute per kilogram of solvent. * Formula: $m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$. * Temperature independent, hence preferred for colligative properties.

4. Applications:

Antifreeze: — Ethylene glycol in car radiators lowers freezing point of coolant.
De-icing: — Salt (NaCl, CaCl$_2$) on roads melts ice by lowering water's freezing point.
Cryoscopy: — Method to determine molar mass of unknown non-volatile solutes.

5. Common Traps & Tips:

Van't Hoff Factor: — Always consider $i$ for electrolytes. It's a frequent source of error.
Units: — Ensure mass of solvent is in kilograms for molality calculation.
Percentage Concentration: — Convert percentage (w/w) to masses of solute and solvent carefully.
Final Temperature: — Remember to subtract $\Delta T_f$ from $T_f^0$ to get the solution's freezing point, which will be a negative value for water.
Comparison: — For equimolal solutions, the one with the highest $i$ will have the lowest freezing point (greatest $\Delta T_f$).

Vyyuha Quick Recall

For Depression, I Know Molality!

Freezing Point
Depression
I — (Van't Hoff factor)
K — (Cryoscopic constant, $K_f$)
Molality ($m$)

This helps recall the formula: $\Delta T_f = i \cdot K_f \cdot m$