Types Of Solutions

In everyday life, we rarely encounter pure substances. Most substances are mixtures of two or more pure components. Their properties depend on their composition. A solution is defined as a homogeneous mixture of two or more components. In a homogeneous mixture, the composition and properties are uniform throughout.

In a solution, the component present in the largest quantity is generally called the solvent. The solvent determines the physical state of the solution (solid, liquid, or gas). The other component(s) present are called solute(s).

We will primarily consider binary solutions, which consist of two components (one solute and one solvent).

Binary solutions can be formed from components in any of the three physical states (solid, liquid, gas), resulting in nine possible types of binary solutions based on the states of solute and solvent:

Type of Solution	Solute	Solvent	Common Examples
Gaseous Solutions	Gas	Gas	Mixture of oxygen and nitrogen gases
	Liquid	Gas	Chloroform mixed with nitrogen gas
	Solid	Gas	Camphor in nitrogen gas
Liquid Solutions	Gas	Liquid	Oxygen dissolved in water (soda water is CO2 in water)
	Liquid	Liquid	Ethanol dissolved in water (Alcoholic drinks)
	Solid	Liquid	Glucose dissolved in water (Sugar solution)
Solid Solutions	Gas	Solid	Solution of hydrogen in palladium
	Liquid	Solid	Amalgam of mercury with sodium (dental fillings)
	Solid	Solid	Copper dissolved in gold (Alloys like brass, bronze)

This unit will focus mainly on liquid solutions where the solvent is a liquid.

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Expressing Concentration Of Solutions

The composition of a solution is described by its concentration. Concentration can be expressed qualitatively (e.g., dilute or concentrated solution) or quantitatively. Quantitative expressions of concentration are necessary for accurate descriptions and calculations.

Several quantitative ways to express the concentration of a solution:

1. Mass percentage (w/w): Mass of a component per 100 parts by mass of the solution.

   $\textsf{Mass % of a component} = \frac{\textsf{Mass of the component in the solution}}{\textsf{Total mass of the solution}} \times 100$

   Example: 10% glucose in water by mass means 10 g glucose in 90 g water (100 g solution). Used in industrial applications (e.g., commercial bleach is 3.62% $\textsf{NaClO}$ by mass).
2. Volume percentage (V/V): Volume of a component per 100 parts by volume of the solution.

   $\textsf{Volume % of a component} = \frac{\textsf{Volume of the component}}{\textsf{Total volume of solution}} \times 100$

   Example: 10% ethanol in water means 10 mL ethanol in water to make a total volume of 100 mL solution. Commonly used for solutions of liquids in liquids (e.g., alcohol percentage in beverages). 35% (v/v) ethylene glycol antifreeze lowers water freezing point to 255.4 K.
3. Mass by volume percentage (w/V): Mass of solute dissolved in 100 mL of the solution. Commonly used in medicine and pharmacy.
4. Parts per million (ppm): Used for very dilute solutions, where the solute is in trace quantities.

   $\textsf{Parts per million (ppm)} = \frac{\textsf{Number of parts of the component}}{\textsf{Total number of parts of all components of the solution}} \times 10^6$

   ppm can be expressed as mass to mass, volume to volume, or mass to volume. Example: 6 $\times$ 10$^{-3}$ g $\textsf{O}_2$ in 1030 g seawater is ~5.8 ppm. Pollutant concentrations (water/air) often in $\mu$g/mL or ppm.
5. Mole fraction (x): Ratio of the number of moles of one component to the total number of moles of all components in the solution.

   For a binary solution with components A and B (moles $\textsf{n}_\textsf{A}$ and $\textsf{n}_\textsf{B}$):

   $\textsf{x}_\textsf{A} = \frac{\textsf{n}_\textsf{A}}{\textsf{n}_\textsf{A} + \textsf{n}_\textsf{B}}$ ; $\textsf{x}_\textsf{B} = \frac{\textsf{n}_\textsf{B}}{\textsf{n}_\textsf{A} + \textsf{n}_\textsf{B}}$

   For a solution with 'i' components, mole fraction of component 'i': $\textsf{x}_\textsf{i} = \frac{\textsf{n}_\textsf{i}}{\Sigma \textsf{n}_\textsf{j}}$. The sum of mole fractions of all components in a solution is always unity: $\textsf{x}_1 + \textsf{x}_2 + ... + \textsf{x}_\textsf{i} = 1$. Useful for relating physical properties (like vapor pressure) and for gas mixtures.

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6. Molarity (M): Number of moles of solute dissolved in one litre (or $\textsf{dm}^3$) of solution.

   $\textsf{Molarity (M)} = \frac{\textsf{Moles of solute}}{\textsf{Volume of solution in litre}}$

   Example: 0.25 mol/L $\textsf{NaOH}$ solution means 0.25 mol $\textsf{NaOH}$ in 1 L solution.

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7. Molality (m): Number of moles of solute per kilogram (kg) of the solvent.

   $\textsf{Molality (m)} = \frac{\textsf{Moles of solute}}{\textsf{Mass of solvent in kg}}$

   Example: 1.00 mol/kg $\textsf{KCl}$ solution means 1 mol $\textsf{KCl}$ (74.5 g) in 1 kg water.

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Each concentration unit has uses. Mass %, ppm, mole fraction, and molality are temperature-independent as they involve masses. Molarity is temperature-dependent because volume changes with temperature.

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Solubility

Solubility is defined as the maximum amount of a substance that can dissolve in a specified amount of solvent at a specified temperature and pressure. It depends on the nature of the solute and solvent, as well as temperature and pressure.

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Solubility Of A Solid In A Liquid

The solubility of a solid in a liquid depends on the nature of both. Generally, like dissolves like: polar solutes dissolve in polar solvents, and nonpolar solutes dissolve in nonpolar solvents (e.g., $\textsf{NaCl}$ and sugar in water, naphthalene and anthracene in benzene).

When a solid solute is added to a solvent, some solute dissolves (dissolution). Simultaneously, solute particles from the solution collide with the undissolved solid and separate out (crystallisation). A state of dynamic equilibrium is reached when the rates of dissolution and crystallisation are equal:

$\textsf{Solute (solid)} + \textsf{Solvent} \rightleftharpoons \textsf{Solution}$

At equilibrium, the solution is saturated – no more solute can dissolve at that temperature and pressure. An unsaturated solution can dissolve more solute. The concentration of solute in a saturated solution is its solubility.

Effect of temperature on solubility of a solid in a liquid: Governed by Le Chatelier's principle.

• If dissolution is endothermic ($\Delta_\text{sol} H > 0$), solubility increases with increasing temperature.
• If dissolution is exothermic ($\Delta_\text{sol} H < 0$), solubility decreases with increasing temperature.

Effect of pressure on solubility of a solid in a liquid: Pressure has no significant effect because solids and liquids are highly incompressible.

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Solubility Of A Gas In A Liquid

Gases also dissolve in liquids (e.g., oxygen in water supporting aquatic life). Solubility of gases in liquids is significantly affected by temperature and pressure.

Effect of pressure: Solubility of gases in liquids increases with increasing pressure. Increasing pressure above the liquid increases the number of gas particles striking the surface and entering the solution, increasing solubility until equilibrium is re-established at a higher concentration.

Henry’s Law: Gives the quantitative relationship between pressure and gas solubility. At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid surface.

Using mole fraction (x) as a measure of solubility, Henry's Law states that the partial pressure of the gas in the vapor phase (p) is proportional to its mole fraction in the solution (x):

$\textsf{p} = \textsf{K}_\textsf{H} \textsf{x}$

Where $\textsf{K}_\textsf{H}$ is Henry's law constant, specific for a given gas and solvent at a specific temperature. A plot of partial pressure vs. mole fraction is linear, with the slope being $\textsf{K}_\textsf{H}$.

Different gases have different $\textsf{K}_\textsf{H}$ values. Higher $\textsf{K}_\textsf{H}$ at a given pressure means lower gas solubility.

Effect of temperature: Solubility of gases in liquids decreases with increasing temperature. Dissolving a gas is an exothermic process (heat is evolved, like condensation). By Le Chatelier's principle, increasing temperature shifts the equilibrium towards the undissolved gas, decreasing solubility. This is why aquatic life thrives better in colder water with higher dissolved oxygen.

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Applications of Henry's Law:

• Soft drinks and soda water are bottled under high $\textsf{CO}_2$ pressure to increase $\textsf{CO}_2$ solubility.
• Scuba divers use tanks filled with air diluted with helium (11.7% He, 56.2% $\textsf{N}_2$, 32.1% $\textsf{O}_2$). Helium is used because it is less soluble in blood than nitrogen at high pressures. Rapid pressure decrease during ascent from deep dives causes dissolved gases to bubble out of blood, leading to 'bends' (painful, dangerous condition). Using He minimises this.
• Low partial pressure of $\textsf{O}_2$ at high altitudes leads to low $\textsf{O}_2$ concentration in blood and tissues (anoxia), causing weakness and impaired thinking in climbers.

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Vapour Pressure Of Liquid Solutions

Liquid solutions are formed when a liquid is the solvent. The solute can be a gas, liquid, or solid. We focus on binary solutions of liquids in liquids and solids in liquids.

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Vapour Pressure Of Liquid-Liquid Solutions

Consider a binary solution of two volatile liquids (components 1 and 2) in a closed vessel. Both liquids evaporate, and an equilibrium is reached between the liquid and vapor phases. The total pressure above the solution is the sum of the partial pressures of the components.

Raoult’s Law (for volatile liquid solutions): States that for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction in the solution.

• For component 1: $\textsf{p}_1 \propto \textsf{x}_1 \implies \textsf{p}_1 = \textsf{p}^0_1 \textsf{x}_1$, where $\textsf{p}_1$ is the partial vapor pressure of component 1, $\textsf{x}_1$ is its mole fraction in the solution, and $\textsf{p}^0_1$ is the vapor pressure of pure component 1 at the same temperature.
• For component 2: $\textsf{p}_2 = \textsf{p}^0_2 \textsf{x}_2$.

According to Dalton's Law of Partial Pressures, the total vapor pressure ($\textsf{p}_\text{total}$) over the solution is the sum of the partial pressures:

$\textsf{p}_\text{total} = \textsf{p}_1 + \textsf{p}_2$

Substituting Raoult's Law: $\textsf{p}_\text{total} = \textsf{p}^0_1 \textsf{x}_1 + \textsf{p}^0_2 \textsf{x}_2$.

Since $\textsf{x}_1 + \textsf{x}_2 = 1$, we can write $\textsf{x}_1 = 1 - \textsf{x}_2$.

$\textsf{p}_\text{total} = \textsf{p}^0_1 (1 - \textsf{x}_2) + \textsf{p}^0_2 \textsf{x}_2 = \textsf{p}^0_1 - \textsf{p}^0_1 \textsf{x}_2 + \textsf{p}^0_2 \textsf{x}_2$

$\textsf{p}_\text{total} = \textsf{p}^0_1 + (\textsf{p}^0_2 - \textsf{p}^0_1) \textsf{x}_2$

This shows that $\textsf{p}_\text{total}$ varies linearly with the mole fraction of either component.

If component 1 is less volatile than component 2 ($\textsf{p}^0_1 < \textsf{p}^0_2$), $\textsf{p}_\text{total}$ increases linearly with increasing $\textsf{x}_2$ (and decreases with increasing $\textsf{x}_1$).

The composition of the vapor phase in equilibrium with the solution is given by the mole fractions in the vapor phase ($\textsf{y}_1, \textsf{y}_2$). Using Dalton's Law:

$\textsf{p}_1 = \textsf{y}_1 \textsf{p}_\text{total}$ ; $\textsf{p}_2 = \textsf{y}_2 \textsf{p}_\text{total}$ ; Generally, $\textsf{p}_\text{i} = \textsf{y}_\text{i} \textsf{p}_\text{total}$.

The vapor phase will be richer in the component that is more volatile (has a higher pure vapor pressure) because its partial pressure in the solution will be higher for the same mole fraction.

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Raoult’S Law As A Special Case Of Henry’S Law

For a solution of a gas in a liquid, Henry's Law states $\textsf{p} = \textsf{K}_\text{H} \textsf{x}$, where p is the partial pressure of the gas above the solution, x is its mole fraction in the solution, and $\textsf{K}_\text{H}$ is Henry's constant.

Raoult's Law states that for a volatile component in a solution, $\textsf{p}_\text{i} = \textsf{p}^0_\text{i} \textsf{x}_\text{i}$.

Comparing the two laws, both state that the partial pressure of a component above the solution is directly proportional to its mole fraction in the solution. The difference lies in the proportionality constant: $\textsf{K}_\text{H}$ for a gas in solution, and $\textsf{p}^0_\text{i}$ for a volatile liquid component in solution.

Thus, Raoult's Law can be considered a special case of Henry's Law where Henry's constant $\textsf{K}_\text{H}$ is equal to the vapor pressure of the pure component $\textsf{p}^0_\text{i}$.

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Vapour Pressure Of Solutions Of Solids In Liquids

This class of solutions has a non-volatile solute dissolved in a volatile liquid solvent. The vapor pressure of the solution comes solely from the solvent.

Adding a non-volatile solute to a pure solvent lowers the vapor pressure of the solvent. This is because the solute particles occupy some area on the surface of the solution, reducing the fraction of the surface available for solvent molecules to evaporate. Consequently, fewer solvent molecules escape into the vapor phase, resulting in lower vapor pressure compared to the pure solvent at the same temperature.

The decrease in vapor pressure depends on the quantity of non-volatile solute, not its nature (e.g., 1 mol sucrose lowers water vapor pressure similarly to 1 mol urea at the same temperature/quantity).

Raoult's Law can be applied to these solutions. Let the solvent be component 1 and the non-volatile solute be component 2.

The vapor pressure of the solvent in the solution ($\textsf{p}_1$) is proportional to its mole fraction ($\textsf{x}_1$):

$\textsf{p}_1 = \textsf{p}^0_1 \textsf{x}_1$

Where $\textsf{p}^0_1$ is the vapor pressure of the pure solvent. A plot of $\textsf{p}_1$ vs. $\textsf{x}_1$ is linear.

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Ideal And Nonideal Solutions

Liquid-liquid solutions are classified based on how well they obey Raoult's Law.

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Ideal Solutions

An ideal solution obeys Raoult's law over the entire range of concentrations and temperatures.

Properties of ideal solutions:

• Enthalpy of mixing is zero: $\Delta_\text{mix} H = 0$ (no heat absorbed or released when components are mixed).
• Volume of mixing is zero: $\Delta_\text{mix} V = 0$ (volume of solution equals the sum of volumes of components).

Molecular explanation: In pure components A and B, intermolecular forces are A-A and B-B. In the solution, A-B interactions also exist. An ideal solution forms when the intermolecular attractive forces between A-A, B-B, and A-B are very similar in magnitude.

Perfectly ideal solutions are rare. Solutions that are nearly ideal include mixtures of n-hexane and n-heptane, bromoethane and chloroethane, benzene and toluene.

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Non-Ideal Solutions

A non-ideal solution does not obey Raoult's law over the entire range of concentration. Their vapor pressure is either higher or lower than predicted by Raoult's law.

• Positive Deviation from Raoult's Law: Vapour pressure is higher than predicted by Raoult's Law ($\textsf{p}_1 > \textsf{p}^0_1 \textsf{x}_1$, $\textsf{p}_2 > \textsf{p}^0_2 \textsf{x}_2$, $\textsf{p}_\text{total} > \textsf{p}^0_1 \textsf{x}_1 + \textsf{p}^0_2 \textsf{x}_2$).
  Molecular explanation: A-B interactions are weaker than A-A and B-B interactions. Molecules can escape from the solution surface more easily, increasing vapor pressure.
  • Example: Ethanol and acetone mixture. Hydrogen bonds in pure ethanol are broken by acetone, weakening interactions.
  • Example: Carbon disulphide and acetone mixture. Dipolar interactions are weaker than in pure components.
  For these solutions, $\Delta_\text{mix} H > 0$ (endothermic mixing) and $\Delta_\text{mix} V > 0$ (volume expands on mixing).
• Negative Deviation from Raoult's Law: Vapour pressure is lower than Raoult's Law prediction ($\textsf{p}_1 < \textsf{p}^0_1 \textsf{x}_1$, $\textsf{p}_2 < \textsf{p}^0_2 \textsf{x}_2$, $\textsf{p}_\text{total} < \textsf{p}^0_1 \textsf{x}_1 + \textsf{p}^0_2 \textsf{x}_2$).
  Molecular explanation: A-B interactions are stronger than A-A and B-B interactions. Molecules are held more tightly in solution, reducing their escaping tendency and lowering vapor pressure.
  • Example: Phenol and aniline mixture (intermolecular hydrogen bonding between phenol and aniline is stronger than in pure components).
  • Example: Chloroform and acetone mixture (hydrogen bonding between chloroform and acetone).

  $\textsf{CHCl}_3 + \textsf{CH}_3\text{COCH}_3 \implies \textsf{CHCl}_3 \cdots \textsf{O=C(CH}_3)_2$ (Hydrogen bond)

  For these solutions, $\Delta_\text{mix} H < 0$ (exothermic mixing) and $\Delta_\text{mix} V < 0$ (volume contracts on mixing).

Azeotropes: Binary mixtures that boil at a constant temperature like a pure liquid and have the same composition in both liquid and vapor phases at that temperature. Components cannot be separated by fractional distillation at the azeotropic composition. Formed by solutions showing large deviations from Raoult's Law.

• Minimum boiling azeotrope: Formed by solutions showing large positive deviations. Boils at a lower temperature than either pure component. Example: Ethanol-water mixture (95% ethanol by volume).
• Maximum boiling azeotrope: Formed by solutions showing large negative deviations. Boils at a higher temperature than either pure component. Example: Nitric acid-water mixture (68% nitric acid by mass, boiling point 393.5 K).

Colligative Properties And Determination Of Molar Mass

Adding a non-volatile solute to a volatile solvent lowers the solvent's vapor pressure. This vapor pressure lowering leads to several other solution properties that depend on the number of solute particles, not their identity or nature. These are called colligative properties.

The four colligative properties are:

1. Relative lowering of vapour pressure of the solvent.
2. Elevation of boiling point of the solvent.
3. Depression of freezing point of the solvent.
4. Osmotic pressure of the solution.

These properties are useful for determining the molar masses of unknown solutes.

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Relative Lowering Of Vapour Pressure

The lowering of vapor pressure when a non-volatile solute is added is $\Delta\textsf{p}_1 = \textsf{p}^0_1 - \textsf{p}_1$, where $\textsf{p}^0_1$ is the vapor pressure of the pure solvent and $\textsf{p}_1$ is the vapor pressure of the solvent in the solution.

According to Raoult's Law for solutions with non-volatile solutes (component 2): $\textsf{p}_1 = \textsf{p}^0_1 \textsf{x}_1$.

So, $\Delta\textsf{p}_1 = \textsf{p}^0_1 - \textsf{p}^0_1 \textsf{x}_1 = \textsf{p}^0_1 (1 - \textsf{x}_1)$.

Since $\textsf{x}_1 + \textsf{x}_2 = 1$, we have $1 - \textsf{x}_1 = \textsf{x}_2$, where $\textsf{x}_2$ is the mole fraction of the solute.

$\Delta\textsf{p}_1 = \textsf{p}^0_1 \textsf{x}_2$.

Rearranging, we get the relative lowering of vapour pressure:

$\frac{\Delta\textsf{p}_1}{\textsf{p}^0_1} = \frac{\textsf{p}^0_1 - \textsf{p}_1}{\textsf{p}^0_1} = \textsf{x}_2$

Thus, the relative lowering of vapor pressure of the solvent is equal to the mole fraction of the non-volatile solute in the solution. For a solution with multiple non-volatile solutes, relative lowering equals the sum of mole fractions of all solutes.

The mole fraction of solute $\textsf{x}_2$ can be expressed in terms of moles $\textsf{n}_1$ (solvent) and $\textsf{n}_2$ (solute): $\textsf{x}_2 = \frac{\textsf{n}_2}{\textsf{n}_1 + \textsf{n}_2}$.

$\frac{\textsf{p}^0_1 - \textsf{p}_1}{\textsf{p}^0_1} = \frac{\textsf{n}_2}{\textsf{n}_1 + \textsf{n}_2}$

For very dilute solutions, $\textsf{n}_2 < < \textsf{n}_1$, so the denominator $\textsf{n}_1 + \textsf{n}_2 \approx \textsf{n}_1$.

$\frac{\textsf{p}^0_1 - \textsf{p}_1}{\textsf{p}^0_1} \approx \frac{\textsf{n}_2}{\textsf{n}_1}$

Moles $\textsf{n}_1 = \frac{\textsf{w}_1}{\textsf{M}_1}$ (mass/molar mass of solvent) and $\textsf{n}_2 = \frac{\textsf{w}_2}{\textsf{M}_2}$ (mass/molar mass of solute).

$\frac{\textsf{p}^0_1 - \textsf{p}_1}{\textsf{p}^0_1} = \frac{\textsf{w}_2/\textsf{M}_2}{\textsf{w}_1/\textsf{M}_1} = \frac{\textsf{w}_2 \textsf{M}_1}{\textsf{w}_1 \textsf{M}_2}$

This equation allows for the calculation of the molar mass of the solute ($\textsf{M}_2$) if other quantities are known.

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Elevation Of Boiling Point

The boiling point of a liquid is the temperature at which its vapor pressure equals the atmospheric pressure. Adding a non-volatile solute lowers the vapor pressure of the solvent. To reach the boiling point (vapor pressure = atmospheric pressure), the solution needs to be heated to a higher temperature than the pure solvent.

Thus, the boiling point of a solution is always higher than the boiling point of the pure solvent. This increase is called the elevation of boiling point ($\Delta\textsf{T}_\text{b}$).

Let $\textsf{T}^0_\text{b}$ be the boiling point of pure solvent and $\textsf{T}_\text{b}$ be the boiling point of the solution. $\Delta\textsf{T}_\text{b} = \textsf{T}_\text{b} - \textsf{T}^0_\text{b}$.

For dilute solutions, the elevation of boiling point is directly proportional to the molality (m) of the solute in the solution:

$\Delta\textsf{T}_\text{b} \propto \textsf{m}$

$\Delta\textsf{T}_\text{b} = \textsf{K}_\text{b} \textsf{m}$

Where $\textsf{K}_\text{b}$ is the Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant). Its unit is K kg mol$^{-1}$. $\textsf{K}_\text{b}$ values are specific for different solvents.

Molality (m) is moles of solute ($\textsf{n}_2$) per kg of solvent ($\textsf{w}_1$ in kg). If $\textsf{w}_2$ g of solute (molar mass $\textsf{M}_2$) is in $\textsf{w}_1$ g of solvent, m = $\frac{\textsf{w}_2/\textsf{M}_2}{\textsf{w}_1/1000} = \frac{\textsf{w}_2 \times 1000}{\textsf{M}_2 \times \textsf{w}_1}$.

$\Delta\textsf{T}_\text{b} = \textsf{K}_\text{b} \frac{\textsf{w}_2 \times 1000}{\textsf{M}_2 \times \textsf{w}_1}$

This equation can be rearranged to find the molar mass of the solute $\textsf{M}_2$:

$\textsf{M}_2 = \frac{\textsf{K}_\text{b} \times \textsf{w}_2 \times 1000}{\Delta\textsf{T}_\text{b} \times \textsf{w}_1}$

$\textsf{K}_\text{b}$ can be calculated using the relation: $\textsf{K}_\text{b} = \frac{\textsf{R} \textsf{M}_1 \textsf{T}^2_\text{b}}{1000 \Delta_\text{vap} H}$ (where $\textsf{M}_1$ is molar mass of solvent, $\textsf{T}_\text{b}$ is boiling point of pure solvent in K, $\Delta_\text{vap} H$ is enthalpy of vaporisation).

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Table 2.3 gives $\textsf{K}_\text{b}$ and $\textsf{K}_\text{f}$ values for common solvents.

Solvent	b.p./K	Kb/K kg mol$^{-1}$	f.p./K	Kf/K kg mol$^{-1}$
Water	373.15	0.52	273.0	1.86
Ethanol	351.5	1.20	155.7	1.99
Cyclohexane	353.74	2.79	279.55	20.00
Benzene	353.3	2.53	278.6	5.12
Chloroform	334.4	3.63	209.6	4.79
Carbon tetrachloride	350.0	5.03	250.5	31.8
Carbon disulphide	319.4	2.34	164.2	3.83
Diethyl ether	307.8	2.02	156.9	1.79
Acetic acid	391.1	2.93	290.0	3.90

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Depression Of Freezing Point

Adding a non-volatile solute to a solvent lowers its vapor pressure. The freezing point of a substance is the temperature where its solid and liquid phases are in dynamic equilibrium, meaning their vapor pressures are equal.

In a solution with a non-volatile solute, the solvent's vapor pressure is lower than that of the pure solid solvent at any given temperature. Thus, the solution's vapor pressure will equal the pure solid solvent's vapor pressure at a lower temperature. This means the freezing point of the solution is lower than that of the pure solvent.

This decrease in freezing point is called the depression of freezing point ($\Delta\textsf{T}_\text{f}$).

Let $\textsf{T}^0_\text{f}$ be the freezing point of pure solvent and $\textsf{T}_\text{f}$ be the freezing point of the solution. $\Delta\textsf{T}_\text{f} = \textsf{T}^0_\text{f} - \textsf{T}_\text{f}$.

For dilute solutions, the depression of freezing point is directly proportional to the molality (m) of the solution:

$\Delta\textsf{T}_\text{f} \propto \textsf{m}$

$\Delta\textsf{T}_\text{f} = \textsf{K}_\text{f} \textsf{m}$

Where $\textsf{K}_\text{f}$ is the Freezing Point Depression Constant or Molal Depression Constant (Cryoscopic Constant). Its unit is K kg mol$^{-1}$. $\textsf{K}_\text{f}$ values are specific for different solvents.

Using the molality expression $m = \frac{\textsf{w}_2 \times 1000}{\textsf{M}_2 \times \textsf{w}_1}$, we get:

$\Delta\textsf{T}_\text{f} = \textsf{K}_\text{f} \frac{\textsf{w}_2 \times 1000}{\textsf{M}_2 \times \textsf{w}_1}$

Rearranging to find the molar mass of the solute $\textsf{M}_2$:

$\textsf{M}_2 = \frac{\textsf{K}_\text{f} \times \textsf{w}_2 \times 1000}{\Delta\textsf{T}_\text{f} \times \textsf{w}_1}$

$\textsf{K}_\text{f}$ can be calculated using the relation: $\textsf{K}_\text{f} = \frac{\textsf{R} \textsf{M}_1 \textsf{T}^2_\text{f}}{1000 \Delta_\text{fus} H}$ (where $\textsf{M}_1$ is molar mass of solvent, $\textsf{T}_\text{f}$ is freezing point of pure solvent in K, $\Delta_\text{fus} H$ is enthalpy of fusion).

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Osmosis And Osmotic Pressure

Many natural phenomena (raw mangoes in brine shrivelling, wilted flowers reviving in water, blood cells collapsing in saline water) involve the movement of solvent across a membrane.

Certain membranes, naturally occurring (pig's bladder, parchment) or synthetic (cellophane), have submicroscopic pores that allow small solvent molecules (like water) to pass through but hinder the passage of larger solute molecules. These are called semipermeable membranes (SPM).

Osmosis: The spontaneous flow of solvent molecules from a region of higher solvent concentration (lower solute concentration, e.g., pure solvent or dilute solution) to a region of lower solvent concentration (higher solute concentration, e.g., concentrated solution) across a semipermeable membrane.

The flow continues until equilibrium is reached. The solvent flows from lower solute concentration to higher solute concentration.

Osmotic pressure ($\pi$ or P): The excess pressure that must be applied to the solution side to stop the flow of solvent across a semipermeable membrane from the pure solvent side (or a more dilute solution side).

Osmotic pressure is a colligative property; it depends on the number of solute particles, not their identity.

For dilute solutions, osmotic pressure is experimentally found to be proportional to the molar concentration (C) of the solution and temperature (T):

$\Pi \propto \textsf{C} \textsf{T}$

$\Pi = \textsf{C} \textsf{R} \textsf{T}$ (where R is the gas constant)

Concentration C is molarity, C = $\textsf{n}_2$/V (moles of solute/volume of solution in litres). So,

$\Pi = \frac{\textsf{n}_2}{\textsf{V}} \textsf{R} \textsf{T}$ or $\Pi \textsf{V} = \textsf{n}_2 \textsf{R} \textsf{T}$

If $\textsf{w}_2$ grams of solute (molar mass $\textsf{M}_2$) is in volume V, then $\textsf{n}_2 = \textsf{w}_2/\textsf{M}_2$.

$\Pi \textsf{V} = \frac{\textsf{w}_2}{\textsf{M}_2} \textsf{R} \textsf{T}$

Rearranging to find the molar mass of the solute $\textsf{M}_2$:

$\textsf{M}_2 = \frac{\textsf{w}_2 \textsf{R} \textsf{T}}{\Pi \textsf{V}}$

The osmotic pressure method is particularly useful for determining molar masses of biomolecules (proteins, polymers) because:

• Measurements are typically done at room temperature, which is suitable for temperature-sensitive biomolecules.
• Molarity is used, which is easier to measure for solutions than molality.
• Osmotic pressure has a large magnitude even for very dilute solutions, allowing accurate measurements for high molecular weight substances.

Solutions with the same osmotic pressure at a given temperature are isotonic solutions. No osmosis occurs between them. Example: 0.9% (mass/volume) $\textsf{NaCl}$ solution ('normal saline') is isotonic with blood cells.

• Hypertonic solution: Higher solute concentration (higher osmotic pressure) than another solution (e.g., >0.9% $\textsf{NaCl}$ compared to blood cells). Water flows out of cells into the hypertonic solution, causing cells to shrink (crenation).
• Hypotonic solution: Lower solute concentration (lower osmotic pressure) than another solution (e.g., <0.9% $\textsf{NaCl}$ compared to blood cells). Water flows into cells from the hypotonic solution, causing cells to swell and potentially burst (lysis).

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Other phenomena explained by osmosis: Water movement from soil to plant roots, preservation of meat by salting (bacteria lose water by osmosis and die), preservation of fruits by sugaring.

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Reverse Osmosis And Water Purification

If an external pressure greater than the osmotic pressure is applied to the solution side of an SPM, the direction of solvent flow can be reversed. Solvent flows from the solution to the pure solvent side. This phenomenon is called reverse osmosis.

Reverse osmosis is used for the desalination of sea water. Applying pressure greater than osmotic pressure to seawater forces pure water molecules through an SPM, leaving the salts and impurities behind. Common SPM used is cellulose acetate, which is permeable to water but impermeable to ions and impurities.

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Abnormal Molar Masses

Colligative properties depend on the number of solute particles. If a solute undergoes dissociation (breaks into ions) or association (molecules combine) in solution, the actual number of particles in solution differs from the number of moles initially dissolved. This leads to colligative properties having values different from those expected based on the initial number of moles, resulting in an abnormal molar mass when calculated from these properties.

• Dissociation: If a solute dissociates (e.g., $\textsf{KCl} \to \textsf{K}^{+} + \textsf{Cl}^{-}$), the number of particles increases. The observed colligative property is higher than expected. The molar mass calculated from the colligative property is lower than the true molar mass (e.g., $\textsf{KCl}$ 74.5 g/mol might appear as 37.25 g/mol).
• Association: If solute molecules associate (e.g., ethanoic acid dimerizing in benzene due to hydrogen bonding), the number of particles decreases. The observed colligative property is lower than expected. The molar mass calculated from the colligative property is higher than the true molar mass (e.g., ethanoic acid dimerizing).

Example of ethanoic acid dimerisation in benzene:

$2 \textsf{CH}_3\text{COOH} \rightleftharpoons (\textsf{CH}_3\text{COOH})_2$

If all ethanoic acid dimerized, the number of particles would be halved, $\Delta\textsf{T}_\text{b}$ or $\Delta\textsf{T}_\text{f}$ would be halved, and the calculated molar mass would be doubled.

Van't Hoff factor (i): Introduced by van't Hoff to account for the extent of dissociation or association. It relates the observed colligative property to the theoretically calculated one.

$\textsf{i} = \frac{\textsf{Normal molar mass}}{\textsf{Abnormal molar mass}} = \frac{\textsf{Observed colligative property}}{\textsf{Calculated colligative property}} = \frac{\textsf{Total number of moles of particles after association/dissociation}}{\textsf{Number of moles of particles before association/dissociation}}$

• For solutes undergoing dissociation, i > 1. (e.g., i $\approx$ 2 for $\textsf{KCl}$ in water at high dilution).
• For solutes undergoing association, i < 1. (e.g., i $\approx$ 0.5 for ethanoic acid dimerizing in benzene).
• For non-electrolyte solutes that neither associate nor dissociate, i = 1.

The equations for colligative properties are modified by including the van't Hoff factor 'i':

• Relative lowering of vapour pressure: $\frac{\Delta\textsf{p}_1}{\textsf{p}^0_1} = \textsf{i} \textsf{x}_2$
• Elevation of Boiling point: $\Delta\textsf{T}_\text{b} = \textsf{i} \textsf{K}_\text{b} \textsf{m}$
• Depression of Freezing point: $\Delta\textsf{T}_\text{f} = \textsf{i} \textsf{K}_\text{f} \textsf{m}$
• Osmotic pressure: $\Pi = \textsf{i} \textsf{C} \textsf{R} \textsf{T} = \textsf{i} \frac{\textsf{n}_2}{\textsf{V}} \textsf{R} \textsf{T}$

For strong electrolytes, i values approach the number of ions per formula unit as the solution becomes very dilute (e.g., i $\to$ 2 for $\textsf{NaCl}$ and $\textsf{KCl}$, i $\to$ 3 for $\textsf{K}_2\textsf{SO}_4$). At higher concentrations, ion pairing reduces the effective number of particles, so i is slightly less than the theoretical value for complete dissociation.

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