Solutions

TYPES OF SOLUTIONS

**Solution** is a homogeneous mixture of two or more components where composition and properties are uniform throughout. The component present in the largest quantity is the **solvent**, which determines the physical state of the solution. Components present in lesser quantity are called **solutes**. In this unit, we focus on **binary solutions** (containing two components).

Solutions are classified based on the physical states of solute and solvent:

**Gaseous Solutions**: Gas dissolved in gas (O₂ and N₂ in air), liquid in gas (chloroform in N₂), solid in gas (camphor in N₂)

**Liquid Solutions**: Gas in liquid (O₂ in water), liquid in liquid (ethanol in water), solid in liquid (glucose in water)

**Solid Solutions**: Gas in solid (H₂ in palladium), liquid in solid (mercury amalgam with sodium), solid in solid (copper in gold)

The most important for board exams are **liquid solutions**, particularly aqueous solutions where water is the solvent.

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EXPRESSING CONCENTRATION OF SOLUTIONS

Concentration describes the amount of solute dissolved in a given quantity of solvent or solution. It can be expressed qualitatively (dilute/concentrated) or quantitatively using various units.

**Mass Percentage (w/w)**

Mass percentage of a component is defined as:

**Mass % = (Mass of component in solution / Total mass of solution) × 100**

*Example*: 10% glucose solution means 10 g glucose dissolved in 90 g water to give 100 g solution. Commonly used in industrial applications (e.g., commercial bleaching solution contains 3.62% sodium hypochlorite).

**Volume Percentage (V/V)**

**Volume % = (Volume of component / Total volume of solution) × 100**

*Example*: 10% ethanol solution means 10 mL ethanol diluted with water to make total volume 100 mL. Used for liquid solutes and solvents. A 35% (v/v) solution of ethylene glycol (antifreeze) lowers water's freezing point to 255.4 K (–17.6°C).

**Mass by Volume Percentage (w/V)**

Mass of solute dissolved in 100 mL of the solution. Commonly used in medicine and pharmacy.

**Parts Per Million (ppm)**

**ppm = (Number of parts of component / Total number of parts) × 10⁶**

Used when solute is present in **trace quantities**. Can be expressed as mass-to-mass, volume-to-volume, or mass-to-volume. *Example*: Sea water contains about 5.8 ppm dissolved oxygen. Pollutant concentrations in water/air are often expressed as mg mL⁻¹ or ppm.

**Mole Fraction (x)**

For a component in a mixture:

**Mole fraction of A = n_A / (n_A + n_B + ... + n_i)**

Where n represents number of moles.

**Key property**: Sum of all mole fractions in a solution equals **unity** (x₁ + x₂ + ... + x_i = 1).

*Worked Example*: 20% ethylene glycol (C₂H₆O₂) by mass in solution. In 100 g solution: 20 g C₂H₆O₂, 80 g water.

Moles of C₂H₆O₂ = 20/62 = 0.322 mol

Moles of H₂O = 80/18 = 4.444 mol

x(C₂H₆O₂) = 0.322/(0.322 + 4.444) = **0.068**

x(H₂O) = 4.444/(0.322 + 4.444) = **0.932**

**Mole fraction is independent of temperature** and useful in relating vapour pressure to concentration and describing gas mixture calculations.

**Molarity (M)**

**Molarity = Moles of solute / Volume of solution in litres**

Unit: mol L⁻¹ or mol dm⁻³

*Example*: 0.25 M NaOH means 0.25 mol NaOH dissolved in 1 litre of solution.

*Worked Example*: 5 g NaOH in 450 mL solution

Moles of NaOH = 5/40 = 0.125 mol

Molarity = 0.125 mol / 0.45 L = **0.278 M**

**Important**: Molarity is **temperature dependent** (volume changes with temperature).

**Molality (m)**

**Molality = Moles of solute / Mass of solvent in kg**

Unit: mol kg⁻¹

*Example*: 1.00 m KCl means 1 mol (74.5 g) KCl dissolved in 1 kg water.

*Worked Example*: 2.5 g ethanoic acid (CH₃COOH) in 75 g benzene

Molar mass of CH₃COOH = 60 g mol⁻¹

Moles = 2.5/60 = 0.0417 mol

Mass of benzene = 75 g = 0.075 kg

Molality = 0.0417/0.075 = **0.556 m**

**Important**: Molality is **temperature independent** (mass does not change with temperature). Useful in colligative property calculations.

**Comparative Summary**

**Independent of temperature**: Mass %, ppm, mole fraction, molality

**Dependent on temperature**: Molarity

**Industrial use**: Mass %, ppm

**Colligative properties**: Molality, mole fraction

**Gas mixtures**: Mole fraction

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SOLUBILITY

**Solubility** is the maximum amount of solute that can dissolve in a specified amount of solvent at a specified temperature and pressure. It depends on:

1. Nature of solute and solvent

2. Temperature

3. Pressure

**Solubility of Solids in Liquids**

**Dissolution-Crystallisation Equilibrium**:

Solute + Solvent ⇌ Solution

When solute is added to solvent, dissolution (solute entering solution) and crystallisation (solute separating out) occur simultaneously. A **dynamic equilibrium** is reached when both processes occur at the same rate. At this point, the concentration remains constant under given conditions.

**Saturated Solution**: Solution in dynamic equilibrium with undissolved solute; contains maximum solute that can dissolve at that temperature/pressure.

**Unsaturated Solution**: More solute can be dissolved at the same temperature/pressure.

**Effect of Temperature on Solid Solubility**:

According to **Le Chatelier's Principle**:

If dissolution is **endothermic** (ΔH_sol > 0), solubility **increases** with temperature rise

If dissolution is **exothermic** (ΔH_sol < 0), solubility **decreases** with temperature rise

Most solid solutes show increasing solubility with temperature (endothermic dissolution).

**Effect of Pressure on Solid Solubility**:

Pressure has **negligible effect** on solubility of solids in liquids because solids and liquids are incompressible.

**Solubility of Gases in Liquids**

When gas dissolves in liquid, dissolution is an **exothermic process** with heat evolution. At equilibrium, the rate of gas molecules entering solution equals rate of molecules leaving solution.

**Effect of Pressure on Gas Solubility**:

When pressure over solution is increased, the number of gas molecules per unit volume increases, and rate of gas molecules striking the solution surface increases. More gas dissolves until new equilibrium is reached. **Gas solubility increases with increasing pressure**.

**Effect of Temperature on Gas Solubility**:

Since dissolution is exothermic, by Le Chatelier's Principle, **gas solubility decreases with increasing temperature**. This explains why aquatic life is more comfortable in cold water (higher dissolved O₂).

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HENRY'S LAW

**Henry's Law Statement**: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution surface.

**Mathematical Expression**:

**p = K_H × x**

Where:

**p** = partial pressure of gas above solution

**x** = mole fraction of gas in solution

**K_H** = Henry's law constant (at given temperature)

**Interpretation**: K_H is the slope of the p vs. x graph. Higher K_H at a given pressure means lower gas solubility in liquid.

**Properties of K_H**:

Varies with **nature of gas** (different gases have different K_H values)

Varies with **temperature** (increases with temperature, indicating lower solubility)

*Example from Table 1.2*: For N₂ at 293 K, K_H = 76.48 kbar; at 303 K, K_H = 88.84 kbar (increased solubility decrease)

**Real-Life Applications**:

1. **Carbonated Beverages**: Bottles sealed under high pressure (2-4 bar) keep CO₂ dissolved. When bottle opens, pressure drops and CO₂ escapes (fizzing).

2. **Scuba Diving - Bends/Decompression Sickness**: Divers breathe air at high pressure underwater. High pressure increases dissolved N₂ and O₂ in blood. When ascending, pressure decreases gradually. If descent is too rapid, dissolved gases form bubbles in blood, blocking capillaries, causing pain and potentially fatal "bends". Prevention: Use air diluted with helium (11.7% He, 56.2% N₂, 32.1% O₂) because He has low solubility and doesn't cause bends.

3. **High Altitude Effects**: At high altitudes, partial pressure of O₂ is low, leading to low dissolved O₂ in blood and tissues. Causes **anoxia** (weakness, mental confusion, potentially fatal).

**Worked Example**: If N₂ gas is bubbled through water at 293 K exerting 0.987 bar partial pressure, how many millimoles dissolve in 1 litre water? (K_H for N₂ = 76.48 kbar)

Using p = K_H × x: x(N₂) = 0.987 bar / 76,480 bar = 1.29 × 10⁻⁵

In 1 L water: n(H₂O) = 1000 g / 18 g mol⁻¹ = 55.5 mol

x(N₂) = n(N₂) / (n(N₂) + 55.5) ≈ n(N₂) / 55.5 = 1.29 × 10⁻⁵

n(N₂) = 1.29 × 10⁻⁵ × 55.5 = 7.16 × 10⁻⁴ mol = **0.716 mmol**

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IDEAL AND NON-IDEAL SOLUTIONS

**Ideal Solution** is a solution that obeys **Raoult's Law** perfectly over entire range of concentration.

**Raoult's Law**: At constant temperature, the partial pressure of a volatile component in a solution is equal to the product of its mole fraction and its vapour pressure in pure state.

**Mathematical Expression**:

For binary solution of A and B:

**p_A = x_A × p_A°**

**p_B = x_B × p_B°**

Where:

**p_A**, **p_B** = partial pressures of A and B in vapour phase

**x_A**, **x_B** = mole fractions of A and B in solution

**p_A°**, **p_B°** = vapour pressures of pure A and B at same temperature

**Total Vapour Pressure**:

**p_total = p_A + p_B = x_A × p_A° + x_B × p_B°**

Since x_A + x_B = 1, we have x_B = 1 - x_A:

**p_total = x_A × p_A° + (1 - x_A) × p_B° = p_B° + x_A(p_A° - p_B°)**

This is a **linear relationship** between total vapour pressure and mole fraction of more volatile component.

**Conditions for Ideal Solutions**:

For solution to be ideal:

1. **Intermolecular interactions** between A-A, B-B, and A-B should be **similar in strength**

2. **Volume change on mixing** should be negligible (V_mix = 0)

3. **Heat change on mixing** should be negligible (ΔH_mix = 0)

4. **No chemical reaction** between components

**Examples of Ideal Solutions**:

Benzene and toluene

Hexane and heptane (hydrocarbons of similar structure)

Chloroform and benzene (if no H-bonding)

**Non-Ideal Solutions**: Solutions that **deviate from Raoult's Law**. Can show **positive deviation** or **negative deviation**.

**Positive Deviation from Raoult's Law**:

**Vapour pressure > calculated** from Raoult's Law

Occurs when **A-B interactions are weaker** than A-A or B-B interactions

**Partial miscibility** (limited solubility); may show liquid-liquid phase separation

*Examples*: Ethanol + hexane, carbon disulfide + acetone, water + ethanol

**Negative Deviation from Raoult's Law**:

**Vapour pressure < calculated** from Raoult's Law

Occurs when **A-B interactions are stronger** than A-A or B-B interactions

*Examples*: Chloroform + acetone (H-bonding between CHCl₃ and C=O), phenol + aniline

**Azeotropes** (Constant Boiling Mixtures):

In positive deviation: Azeotropic point at **maximum vapour pressure** (minimum boiling point)

In negative deviation: Azeotropic point at **minimum vapour pressure** (maximum boiling point)

At azeotropic composition, vapour has **same composition as liquid**; cannot be separated by fractional distillation

*Example*: Ethanol-water azeotrope (95.6% ethanol, 4.4% water) boils at 78.15°C; cannot obtain pure ethanol by distillation

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COLLIGATIVE PROPERTIES

**Colligative Properties** are properties of solutions that depend **only on the number of solute particles** dissolved in solvent, regardless of the nature or identity of solute particles. These are independent of:

Chemical identity of solute

Chemical identity of solvent (to some extent)

Chemical interactions between solute and solvent

**Four Main Colligative Properties**:

1. **Relative Lowering of Vapour Pressure**

2. **Elevation of Boiling Point**

3. **Depression of Freezing Point**

4. **Osmotic Pressure**

These properties arise because solute particles disrupt solvent-solvent interactions, affecting equilibria involving solvent molecules.

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RELATIVE LOWERING OF VAPOUR PRESSURE

When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solution **decreases compared to pure solvent**.

**Derivation from Raoult's Law**:

For solution with non-volatile solute (solute vapour pressure ≈ 0):

**p_solution = x_solvent × p_solvent° = (n_solvent)/(n_solvent + n_solute) × p_solvent°**

For dilute solutions where n_solute << n_solvent:

**p_solution / p_solvent° ≈ n_solvent / (n_solvent + n_solute) = 1 - (n_solute)/(n_solvent + n_solute)**

**Relative lowering of vapour pressure**:

**(p_solvent° - p_solution) / p_solvent° = (n_solute) / (n_solvent + n_solute) ≈ (n_solute) / (n_solvent)**

Or in terms of **mole fraction of solute**:

**Δp / p° = x_solute**

Where:

Δp = p° - p_solution (lowering of vapour pressure)

p° = vapour pressure of pure solvent

x_solute = mole fraction of solute

**Key Point**: The relative lowering of vapour pressure equals the mole fraction of solute in solution. This is independent of solute's identity—only its number of particles matters.

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ELEVATION OF BOILING POINT (EBULLIOSCOPIC EFFECT)

When non-volatile solute is dissolved in solvent, the **boiling point of solution increases** compared to pure solvent.

**Reason**: At any temperature, vapour pressure of solution is lower than pure solvent (due to lowering of vapour pressure). To restore vapour pressure to atmospheric pressure (needed for boiling), temperature must be increased.

**Mathematical Relationship**:

**ΔT_b = K_b × m**

Where:

**ΔT_b** = elevation of boiling point = T_b(solution) - T_b(solvent)

**K_b** = ebullioscopic constant (boiling point elevation constant) of solvent (°C kg mol⁻¹)

**m** = molality of solute (mol kg⁻¹)

**Derivation** (using thermodynamic approach):

At boiling point equilibrium: ΔG = 0, so ΔH = TΔS

For phase transition: ΔH_vap = T_b × ΔS_vap

For solution: ΔT_b = (R × T_b² × m) / (1000 × ΔH_vap)

Where ΔH_vap is molar heat of vaporization.

Defining K_b = (R × T_b²) / (1000 × ΔH_vap):

**ΔT_b = K_b × m**

**Values of K_b for Common Solvents** (at their normal boiling points):

Water: **0.512 K kg mol⁻¹**

Benzene: **2.53 K kg mol⁻¹**

Chloroform: **3.63 K kg mol⁻¹**

Acetic acid: **3.07 K kg mol⁻¹**

**Important Points**:

1. ΔT_b is **independent of solute identity** (depends only on number of particles)

2. Boiling point elevation is a **colligative property**

3. Used to determine **molar mass** of solute

4. More pronounced in solvents with large K_b values

**Worked Example**: Calculate boiling point elevation for 1 m NaCl solution in water (K_b = 0.512 K kg mol⁻¹, but NaCl gives 2 particles per formula unit; total moles of particles = 2 × 1 = 2 m)

ΔT_b = 0.512 × 2 = **1.024 K**

Boiling point of solution = 373.15 K + 1.024 K = **374.17 K**

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DEPRESSION OF FREEZING POINT (CRYOSCOPIC EFFECT)

When non-volatile solute is dissolved in solvent, the **freezing point of solution decreases** compared to pure solvent.

**Reason**: At any temperature below freezing point of pure solvent, pure solid solvent is in equilibrium with solution. The presence of solute particles in solution **lowers the chemical potential of solvent in liquid phase**, requiring lower temperature to restore equilibrium with solid phase.

**Mathematical Relationship**:

**ΔT_f = K_f × m**

Where:

**ΔT_f** = depression of freezing point = T_f(solvent) - T_f(solution)

**K_f** = cryoscopic constant (freezing point depression constant) of solvent (°C kg mol⁻¹ or K kg mol⁻¹)

**m** = molality of solute (mol kg⁻¹)

**Derivation** (parallel to boiling point elevation):

At freezing point equilibrium: ΔH_fus = T_f × ΔS_fus

For solution: ΔT_f = (R × T_f² × m) / (1000 × ΔH_fus)

Defining K_f = (R × T_f²) / (1000 × ΔH_fus):

**ΔT_f = K_f × m**

**Values of K_f for Common Solvents** (at their normal freezing points):

Water: **1.86 K kg mol⁻¹**

Benzene: **5.12 K kg mol⁻¹**

Chloroform: **4.68 K kg mol⁻¹**

Acetic acid: **3.9 K kg mol⁻¹**

**Important Points**:

1. ΔT_f is **independent of solute identity** (depends only on number of particles)

2. Freezing point depression is a **colligative property**

3. Magnitude of ΔT_f typically **larger than ΔT_b** for same solute and solvent

4. Used to determine **molar mass** of solute

5. More pronounced in solvents with large K_f values

**Real-Life Applications**:

**Antifreeze in cars**: 35% (v/v) ethylene glycol lowers freezing point of water to **255.4 K (−17.6°C)** and raises boiling point

**De-icing of roads**: Salt (NaCl) lowers freezing point of water, allowing roads to remain liquid

**Biological systems**: Presence of dissolved ions/proteins in blood prevents freezing at body temperature and regulates osmotic balance

**Worked Example**: Calculate freezing point depression for 2 m glucose solution in water (K_f = 1.86 K kg mol⁻¹; glucose is non-electrolyte)

ΔT_f = 1.86 × 2 = **3.72 K**

Freezing point of solution = 273.15 K − 3.72 K = **269.43 K** or **−3.72°C**

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OSMOTIC PRESSURE

**Osmosis** is the spontaneous flow of solvent molecules across a **semipermeable membrane** from a region of lower solute concentration (higher solvent concentration) to a region of higher solute concentration (lower solvent concentration).

**Semipermeable Membrane**: A membrane that allows passage of solvent molecules but **blocks passage of solute particles** (either because solute molecules are too large or due to membrane pore characteristics).

**Osmotic Pressure (π)**: The pressure that must be applied to a solution to prevent osmotic flow of solvent into solution, or equivalently, the pressure difference that drives osmosis.

**Mathematical Expression** (van't Hoff Equation):

**π = CRT**

Or equivalently:

**π = (n/V) × RT = (n × RT) / V**

Where:

**π** = osmotic pressure (in atm, bar, or Pa)

**C** = molar concentration of solute (mol L⁻¹)

**n** = number of moles of solute

**V** = volume of solution (in litres)

**R** = gas constant = 0.0821 L atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹

**T** = absolute temperature (K)

**Comparison with Ideal Gas Law**:

The van't Hoff equation for osmotic pressure has **identical form** to ideal gas law (PV = nRT). This remarkable similarity suggests **solute particles behave like gas molecules** in distributing throughout solution, exerting pressure on semipermeable membrane.

**Derivation** (Thermodynamic basis):

For osmotic equilibrium across semipermeable membrane:

Chemical potential of solvent in pure solvent = chemical potential of solvent in solution

This leads to pressure difference between solution and pure solvent side

At equilibrium: π = (RT/V_m) × ln(1 / (1 - x_solute)) ≈ (RT/V_m) × x_solute (for dilute solutions)

Where V_m is molar volume of solvent

For dilute solutions: π = CRT

**Key Points about Osmotic Pressure**:

1. **Colligative property**: Depends only on number of solute particles, not their identity

2. **Largest effect** among colligative properties: Osmotic pressure is typically 10-100 times larger than freezing point depression or boiling point elevation (e.g., 1 M NaCl solution at 298 K: π = 2 × 298 × 0.0821 ≈ 49 atm)

3. **Temperature dependent**: π increases proportionally with absolute temperature

4. **Independent of solute nature**: Same concentration of any solute produces same osmotic pressure

5. **Most easily measured** of colligative properties (modern osmometers measure π directly)

**Applications**:

1. **Biological Systems**:

Blood osmotic pressure ≈ 7.8 atm at body temperature (98.6°F)

Intravenous (IV) fluids must have osmotic pressure equal to blood plasma (isotonic solutions)

**Hypertonic solutions** (π > blood π): Water flows out of cells, cells shrivel (crenation)

**Hypotonic solutions** (π < blood π): Water flows into cells, cells swell and burst (hemolysis)

**Isotonic solutions** (π = blood π): No net water flow; maintains cell integrity

2. **Plant Cells**: Turgor pressure (osmotic pressure) maintains cell rigidity and plant structure

3. **Desalination**: Reverse osmosis uses pressure > π to drive water through semipermeable membrane, leaving salt behind

4. **Food Preservation**: High sugar/salt concentrations create hypertonic environment, drawing water from microorganisms and preventing growth

**Worked Example**: Calculate osmotic pressure of 0.5 M glucose solution at 298 K

π = CRT = 0.5 mol L⁻¹ × 0.0821 L atm K⁻¹ mol⁻¹ × 298 K

π = **12.2 atm**

**Worked Example**: A 0.1 M NaCl solution at 298 K (Note: NaCl dissociates into Na⁺ and Cl⁻)

Effective concentration = 0.1 × 2 = 0.2 M (ions)

π = 0.2 × 0.0821 × 298 = **4.88 atm**

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ABNORMAL COLLIGATIVE PROPERTIES

Some solutes show **colligative properties differing from theoretical predictions**. The discrepancy arises because colligative property equations assume **one solute particle per formula unit**, but some solutes dissociate into multiple particles or associate into larger units.

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

To account for deviation from ideality due to particle number changes, the **van't Hoff factor** is introduced:

**i = (Number of particles produced by 1 formula unit of solute) / 1**

Or experimentally:

**i = (Observed colligative property) / (Theoretical colligative property for 1 particle)**

**Modified Equations for Colligative Properties**:

1. **Relative lowering of vapour pressure**: Δp/p° = i × x_solute

2. **Elevation of boiling point**: ΔT_b = i × K_b × m

3. **Depression of freezing point**: ΔT_f = i × K_f × m

4. **Osmotic pressure**: π = i × CRT

**Cases of Abnormal i Values**:

**Case 1: Electrolytes (Strong Dissociation)**

Strong electrolytes completely dissociate in solution, producing i > 1.

**NaCl**: i = 2 (dissociates as NaCl → Na⁺ + Cl⁻)

**CaCl₂**: i = 3 (dissociates as CaCl₂ → Ca²⁺ + 2Cl⁻)

**K₂SO₄**: i = 3 (dissociates as K₂SO₄ → 2K⁺ + SO₄²⁻)

**Glucose** (non-electrolyte): i = 1 (no dissociation)

*Example*: For 0.1 M NaCl, freezing point depression:

ΔT_f = i × K_f × m = 2 × 1.86 × 0.1 = **0.372 K** (not 0.186 K as predicted without considering dissociation)

**Case 2: Weak Electrolytes (Partial Dissociation)**

Weak electrolytes partially dissociate, producing **1 < i < theoretical maximum**.

**Acetic acid (CH₃COOH)**: Partially dissociates, i ≈ 1.3−1.4 (not 1, not 2)

**Weak acids/bases**: i between 1 and complete dissociation value

The degree of dissociation (α) determines i:

**i = 1 + (n − 1) × α**

Where n = number of particles produced upon complete dissociation, α = fraction of solute that dissociates

*Example*: If 30% of acetic acid dissociates (α = 0.3):

i = 1 + (2 − 1) × 0.3 = 1 + 0.3 = **1.3**

**Case 3: Association (Solute Particle Aggregation)**

In some solvents, solute molecules associate/aggregate to form larger units, producing **i < 1**.

**Acetic acid in benzene**: Forms dimers (CH₃COOH)₂ through hydrogen bonding

**Benzoic acid in certain solvents**: Similarly dimerizes

Association reduces number of particles in solution, so observed colligative properties are smaller than expected for monomeric solute.

**Relationship between i and Molar Mass**:

The van't Hoff factor can be determined experimentally using colligative properties:

From ΔT_f = i × K_f × m and **m = (mass of solute in grams) / (molar mass in g mol⁻¹) / (kg of solvent)**:

**Molar mass = (i × K_f × mass of solute × 1000) / (ΔT_f × mass of solvent)**

If i is not accounted for, calculated molar mass will be **anomalous** (abnormally high or low).

**Worked Example**: A 1.00 g sample of NaCl dissolved in 100 g water lowers freezing point by 0.372 K. Calculate apparent molar mass.

Using ΔT_f = i × K_f × m:

0.372 = i × 1.86 × m

Without considering dissociation (i = 1): m = 0.372/1.86 = 0.20 m

Apparent molar mass = 1.00 g / 0.20 m / 0.1 kg = **50 g mol⁻¹** (not 58.5 g mol⁻¹ actual)

**Reason for discrepancy**: NaCl dissociates into 2 ions, so actual i = 2, and true molar mass = 50 × 2 = **100 g mol⁻¹** (approximately correct, accounting for some ionic interactions).

If we use corrected i = 2:

m = 0.372 / (2 × 1.86) = 0.10 m

Molar mass = 1.00 / 0.10 / 0.1 = **100 g mol⁻¹** ≈ 58.5 g mol⁻¹ (discrepancy due to ion pairing)

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DETERMINATION OF MOLAR MASS USING COLLIGATIVE PROPERTIES

Molar mass can be determined experimentally using any colligative property. The choice depends on practical considerations.

**From Freezing Point Depression** (most commonly used):

**Molar mass (M) = (K_f × w_B × 1000) / (w_A × ΔT_f)**

Where:

K_f = c