Colligative Properties Elevation in B.P., Depression in F.P., Relative Lowering

What are Colligative Properties?

The colligative properties of a solution are those properties that depend primarily on the number of solute particles dissolved in a given amount of solvent and not on the chemical identity of the solute. In other words, these properties change with the concentration (amount) of the solute particles, whether they are atoms, molecules, or ions.

The four main colligative properties

• Elevation of boiling point
• Depression of freezing point
• Lowering (and relative lowering) of vapour pressure
• Osmotic pressure

Colligative Properties and Its Types

Applications of colligative properties

Colligative properties are useful in many practical areas such as the pharmaceutical industry, food technology, and chemical engineering. They are exploited to determine molar masses (for non-volatile, non-electrolytic solutes) using methods based on freezing point depression, boiling point elevation, vapour pressure lowering or osmotic pressure.

Note: Colligative relations are derived for ideal dilute solutions. Deviations occur at high concentrations or when strong solute-solvent interactions, association or dissociation occur.

1. Boiling Point Elevation

Boiling point elevation is the increase in the boiling point of a solvent when a non-volatile solute is dissolved in it. A solution boils at a higher temperature than the pure solvent because the vapour pressure of the solvent is lowered by the solute; the temperature must be raised further to make the vapour pressure equal to the external (atmospheric) pressure.

• Boiling point elevation is a colligative property: it depends on the amount (concentration) of solute particles, not their identity.
• The greater the concentration (molality) of solute particles, the larger the elevation in boiling point.

Why boiling point is elevated

Addition of a non-volatile solute lowers the vapour pressure of the solvent. Since boiling occurs when vapour pressure equals external pressure, a lower vapour pressure requires a higher temperature to reach the same vapour pressure. Thus the solution's boiling point increases.

If T_b˚ is the boiling point of the solvent and T_b is the boiling point of the solution.

Elevation in boiling point

Elevation in boiling point (ΔT_b) is defined as the difference between the boiling point of the solution and the boiling point of the pure solvent.

ΔT_b = T_b (solution) - T_b^˚ (pure solvent)

Experimentally for dilute solutions, ΔT_b is proportional to the molality (m) of the solute:

\( \Delta T_{b} \propto m \)

\( \Delta T_{b} = K_{b}\,m \)

K_b is the molal boiling point elevation constant (ebullioscopic constant) of the solvent and has units K·kg·mol^-1.

Molality (m)

Molality (m) = Number of moles of solute / Mass of solvent (in kg)

w = mass of solute in grams
M = molecular wt of solute
W = mass of solvent in gram.)

Significance of Kb

If m = 1 mol kg^-1, then ΔT_b = K_b. Thus K_b equals the elevation in boiling point when 1 mole of solute is dissolved in 1 kg of solvent.

Example 1. A 5 percent aqueous solution by mass of a non-volatile solute boils at 100.15°C. Calculate the molar mass of the solute. Given K_b = 0.52 K·kg·mol^-1.

Solution.

\( \Delta T_b = 100.15^\circ\text{C} - 100.00^\circ\text{C} = 0.15\ \text{K} \)

\( m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.15}{0.52} = 0.2885\ \text{mol kg}^{-1} \)

For 5% w/w solution, 5 g solute in 100 g solution ⇒ solvent = 95 g = 0.095 kg

\( \text{Molality } m = \dfrac{\text{moles of solute}}{\text{kg solvent}} = \dfrac{5/M}{0.095} = 0.2885 \)

\( \dfrac{5}{0.095\,M} = 0.2885 \)

\( M = \dfrac{5}{0.095 \times 0.2885} = 182.4\ \text{g mol}^{-1} \)

Example 2. A solution of 12.5 g urea in 170 g of water gave a boiling point elevation of 0.63 K. Calculate the molar mass of urea. Take K_b = 0.52 K·m^-1.

Solution.

Mass of solute w₂ = 12.5 g; mass of solvent w₁ = 170 g = 0.170 kg

\( \Delta T_b = 0.63\ \text{K} \)

\( m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.63}{0.52} = 1.2115\ \text{mol kg}^{-1} \)

\( \text{moles of solute } n = m \times \text{kg solvent} = 1.2115 \times 0.170 = 0.20596\ \text{mol} \)

\( M = \dfrac{\text{mass}}{n} = \dfrac{12.5}{0.20596} = 60.7\ \text{g mol}^{-1} \)

Example 3. Calculate the molecular mass of a substance, 1.0 g of which on being dissolved in 100 g of solvent gave an elevation of 0.307 K. K_b = 1.84 K·m^-1.

Solution.

Mass of solute = 1.0 g; mass of solvent = 100 g = 0.1 kg

\( \Delta T_b = 0.307\ \text{K} \)

\( m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.307}{1.84} = 0.1668\ \text{mol kg}^{-1} \)

\( \text{moles of solute } = m \times \text{kg solvent} = 0.1668 \times 0.1 = 0.01668\ \text{mol} \)

\( M = \dfrac{1.0\ \text{g}}{0.01668\ \text{mol}} = 59.9\ \text{g mol}^{-1} \)

TRY YOURSELF!

2. Freezing Point Depression

The freezing point of a substance is the temperature at which its liquid and solid phases have the same vapour pressure. When a non-volatile solute is added to a solvent, the freezing point of the resulting solution is lower than that of the pure solvent. This is called freezing point depression.

For dilute solutions, the depression in freezing point is proportional to the molality of the solute:

\( \Delta T_{f} = K_{f}\,m \)

Where:

• ΔT_f is the freezing point depression,
• K_f is the cryoscopic constant (molal freezing point depression constant) of the solvent,
• m is the molality of the solute (mol kg^-1).

Depression in Freezing Point

Why freezing point is depressed

1. At the freezing point there is equilibrium between liquid and solid phases (vapour pressures equal).
2. Addition of a non-volatile solute lowers the vapour pressure of the liquid solution compared with the pure liquid.
3. Because the solution has a lower vapour pressure, the liquid-solid equilibrium is achieved at a lower temperature; hence freezing point decreases.

Expressed as a difference:

ΔT_f = T_f^˚ (pure solvent) - T_f (solution)

Significance of Kf

If m = 1 mol kg^-1, then ΔT_f = K_f. Thus K_f equals the depression in freezing point when 1 mole of solute is dissolved in 1 kg of the solvent. Substances with large K_f are particularly useful for molecular mass determinations (camphor is a common example because of its large K_f).
The values of molal elevation constant (K_b) for some common solvents are given in Table.

Relations involving Kb and Kf

K_b and K_f can be related to thermodynamic quantities:

Where, T_b = boiling point of the solvent,
T_f = freezing point of the solvent,
M_solvent = molar mass of the solvent,
Δ_vapH = enthalpy of vaporization and
Δ_fusH = enthalpy of fusion.

Key concept: Camphor is commonly used in determining the molecular mass of a solute because of its very high cryoscopic constant (K_f).

Example 1. An aqueous solution of a non-volatile solute boils at 100.17°C. At what temperature would it freeze? (K_b = 0.52 K·kg·mol^-1 and K_f = 1.88 K·kg·mol^-1.)

Solution.

\( \Delta T_b = 0.17\ \text{K} \)

\( m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.17}{0.52} = 0.3269\ \text{mol kg}^{-1} \)

\( \Delta T_f = K_f \times m = 1.88 \times 0.3269 = 0.614\ \text{K} \)

Freezing point = 0°C - 0.614 K = -0.614°C (approx).

Example 2. 1.4 g of acetone dissolved in 100 g of benzene gave a solution that froze at 277.12 K. Pure benzene freezes at 278.40 K. 2.8 g of a solid A dissolved in 100 g of benzene gave a solution freezing at 277.76 K. Calculate the molecular weight of A.

Solution.

For acetone + benzene:

\( \Delta T = 278.40 - 277.12 = 1.28\ \text{K} \)

\( \Delta T = \dfrac{1000 K_f w}{M W} \) (use standard molality-based relation)

\( 1.28 = 1000 \times K_f \times \dfrac{1.4}{100 \times 58} \) (equation i)

For solid A + benzene:

\( \Delta T = 278.40 - 277.76 = 0.64\ \text{K} \)

\( 0.64 = 1000 \times K_f \times \dfrac{2.8}{100 \times M} \) (equation ii)

Divide (i) by (ii) and solve for M to obtain M = 232 g mol^-1.

Example 3. Ethylene glycol (HO-CH2-CH2-OH) is used as antifreeze. How much ethylene glycol should be added to 1 kg of water to prevent freezing at -10°C? K_f (water) = 1.86 K·kg·mol^-1. M = 62 g mol^-1.

Solution.

\( \Delta T_f = 10\ \text{K} \)

\( m = \dfrac{\Delta T_f}{K_f} = \dfrac{10}{1.86} = 5.376\ \text{mol kg}^{-1} \)

Moles required = m × kg solvent = 5.376 × 1 = 5.376 mol

Mass = moles × M = 5.376 × 62 = 333.3 g

Example 4. Normal freezing point of a solvent is 150°C. A 0.5 molal solution of urea causes freezing point depression of 2°C. Calculate K_f.

Solution.

\( \Delta T_f = K_f m \)

\( K_f = \frac{\Delta T_f}{m} = \frac{2}{0.5} = 4\ \text{K\,kg\,mol}^{-1} \)

Example 5. A solution of urea in water freezes at 0.400°C. What is the boiling point of the same solution if K_f = 1.86 and K_b = 0.512 K·kg·mol^-1?

Solution.

Let n₂ be moles of solute and w₁ mass of solvent (kg).

\( \Delta T_b = K_b \dfrac{n_2}{w_1} \)

\( \Delta T_f = K_f \dfrac{n_2}{w_1} \)

Divide the two equations:

\( \dfrac{\Delta T_b}{\Delta T_f} = \dfrac{K_b}{K_f} \)

\( \Delta T_b = \Delta T_f \times \dfrac{K_b}{K_f} \)

\( \Delta T_f = 100.00^\circ\text{C} - 0.400^\circ\text{C} = 0.400^\circ\text{C} \) (note: freezing point of pure water is 0.00°C; here the solution freezes at 0.400°C, implying ΔT_f = 0.400)

\( \Delta T_b = 0.400 \times \dfrac{0.512}{1.86} = 0.110\ \text{K} \)

Boiling point of solution = 100.00 + 0.110 = 100.11°C

TRY YOURSELF!

3. Relative Lowering of Vapour Pressure

Relative lowering of vapour pressure is a colligative property: when a non-volatile solute is added to a volatile solvent, the vapour pressure of the solvent above the solution is lower than the vapour pressure of the pure solvent.

Lowering of Vapour Pressure

• The vapour pressure of the solvent above a solution is reduced when a non-volatile solute is present.
• If p^˚ is the vapour pressure of the pure solvent and p is the vapour pressure of the solvent above the solution, the lowering is p^˚ - p.
• The relative lowering of vapour pressure is defined as (p^˚ - p) / p^˚.
• Raoult's law relates vapour pressure of the solvent above a dilute solution to its mole fraction: the vapour pressure of the solvent over the solution equals the mole fraction of the solvent multiplied by the vapour pressure of the pure solvent.
  Here, n is the moles of the solute dissolved in N moles of the solvent.

Raoult's law (ideal dilute solutions)

For a solvent in a solution:

\( p_{\text{solution}} = X_{\text{solvent}} \times p_{\text{solvent}}^{\circ} \)

Therefore, lowering of vapour pressure:

\( p^{\circ} - p = p^{\circ}(1 - X_{\text{solvent}}) = p^{\circ} X_{\text{solute}} \)

So, the relative lowering is

\( \dfrac{p^{\circ} - p}{p^{\circ}} = X_{\text{solute}} \)

Vapour pressure lowering and electrolytes

Relative lowering is a colligative property because it depends on the number of solute particles. Ionic solutes that dissociate produce more particles per formula unit and therefore produce a larger lowering for the same molar concentration than non-electrolytes.
For example, 1 M NaCl (which largely dissociates into Na⁺ and Cl^-) produces roughly twice the effect of 1 M glucose, which remains as single molecules.NaCl(s)→Na+(aq)+Cl–(aq) ⇒ 2 dissolved particles
C₆H₁₂O₆(s)→C₆H₁₂O₆(aq) ⇒1dissolved particles

Example 1: At 25°C the vapour pressure of pure benzene is 93.9 torr. When a non-volatile solute is added, the vapour pressure of benzene is lowered to 91.5 torr. Calculate the mole fraction of the solute and of the solvent.

Solution.

\( p^{\circ} = 93.9\ \text{torr},\quad p = 91.5\ \text{torr} \)

\( \Delta p = p^{\circ} - p = 93.9 - 91.5 = 2.4\ \text{torr} \)

\( X_{\text{solute}} = \dfrac{\Delta p}{p^{\circ}} = \dfrac{2.4}{93.9} = 0.02557 \)

\( X_{\text{solvent}} = 1 - X_{\text{solute}} = 0.97443 \)

Example 2: The vapour pressure of a pure liquid at 298 K is 4.0×10⁴ N m^-2. Adding a non-volatile solute reduces the vapour pressure to 3.65×10⁴ N m^-2. Calculate the relative vapour pressure, lowering and relative lowering.

Solution.

\( p^{\circ} = 4.0\times10^{4}\ \text{N m}^{-2},\quad p = 3.65\times10^{4}\ \text{N m}^{-2} \)

\( \text{Lowering } = p^{\circ} - p = 0.35\times10^{4}\ \text{N m}^{-2} \)

\( \text{Relative vapour pressure } = \dfrac{p}{p^{\circ}} = \dfrac{3.65}{4.0} = 0.9125 \)

\( \text{Relative lowering } = \dfrac{p^{\circ} - p}{p^{\circ}} = 1 - 0.9125 = 0.0875 \)

Example 3: The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g cm^-3. Calculate the vapour pressure of water above this solution. Given: p^˚ (water) = 23.79 mm Hg. Assume complete dissociation.

Solution.

0.438 mol K₂CrO₄ per litre of solution ⇒ mass of solute per L = 0.438 × 194 = 84.972 g

Mass of 1 L solution = 1000 mL × 1.063 g mL^-1 = 1063 g

Mass of water in 1 L solution = 1063 - 84.972 = 978.028 g

Moles of water = 978.028 / 18 = 54.335 mol

Assuming complete dissociation: each K₂CrO₄ → 3 ions ⇒ total solute particles = 3 × 0.438 = 1.314 mol

Mole fraction of water = 54.335 / (54.335 + 1.314) = 0.976

Vapour pressure above solution = X_water × p^˚ = 0.976 × 23.79 = 23.22 mm Hg

Determination of molar masses from lowering of vapour pressure

• By measuring the lowering of vapour pressure produced by dissolving a known mass of a non-volatile solute in a solvent, the molar mass of the solute can be determined (for dilute, non-electrolytic solutes).
• If w gram of solute (molar mass m) is dissolved in W gram of solvent (molar mass M), the relative lowering equals mole fraction of solute. Using relations between moles and masses gives the formula for solving m.

Osmosis and Osmotic Pressure 

Osmosis is a fundamental biological and chemical process in which solvent molecules, typically water, move through a semipermeable membrane from a region of low solute concentration to a region of high solute concentration. This movement aims to balance the concentrations on either side of the membrane.

Semipermeable Membranes (SPM) are critical components in processes like osmosis. They selectively allow certain molecules to pass through while blocking others based on size or other properties. In the context of solutions, they typically allow the solvent (e.g., water) to pass through but prevent the solute (e.g., salt) from crossing. This selective permeability is due to the presence of tiny pores in the membrane.

Key Characteristics of Semipermeable Membranes:

(i) Selective Permeability

• Solvent Passage: The pores are small enough to let solvent molecules pass through.
• Solute Exclusion: Larger solute molecules are blocked due to their size.

(ii) Material and Structure

• Made of materials like cellulose acetate, polyamide, or other polymers.
• The structure consists of a network of submicroscopic pores that act as a sieve.

(iii) Applications

• Biological Systems: Cell membranes, which regulate the passage of nutrients and waste.
• Industrial Uses: Water purification, dialysis, and other filtration processes.

Osmotic Pressure (π)

Osmotic Pressure is a fundamental concept in the study of solutions and osmosis. It refers to the pressure required to prevent the flow of a solvent through a semipermeable membrane into a solution containing a higher concentration of solute.

Role in Osmosis:

When two solutions of different concentrations are separated by an SPM, solvent molecules naturally move from the area of lower solute concentration to the area of higher solute concentration. Osmotic pressure is the force that balances this movement, preventing further solvent flow when equilibrium is reached.

Van't Hoff Laws of Osmotic Pressure

1. Van't Hoff-Boyle's Law:

   • Statement: The osmotic pressure (π) of a solution is directly proportional to its molar concentration (C), when the temperature is kept constant.
   • Mathematical Representation: $\pi \propto C$
   • This means that if the concentration of solute in a solution increases, the osmotic pressure also increases, assuming the temperature does not change.

2. Van't Hoff-Charles' Law:

   • Statement: At constant concentration, the osmotic pressure of a dilute solution is directly proportional to its absolute temperature (T).
   • Mathematical Representation: $\pi \propto T$
   • This implies that if the temperature of a solution rises, the osmotic pressure will also rise, given that the concentration remains the same.

3. Combined Van't Hoff Law:

   • By combining the above two laws, the osmotic pressure is given by: $\pi \propto C\cdot T$⋅T Or, more precisely: $\pi =CRT$
   • Where:
     • $\pi$π is the osmotic pressure.
     • $C$C is the molar concentration of the solution.
     • $R$R is the universal gas constant.
     • $T$T is the absolute temperature in Kelvin.

Isotonic or Iso-Osmotic Solutions

• Definition: Solutions that have the same osmotic pressure at a given temperature are called isotonic or iso-osmotic solutions.
• Behavior: When isotonic solutions are separated by a semipermeable membrane, no net osmosis occurs between them. This means that there will be no movement of water across the membrane since the osmotic pressures are balanced.

Example: Osmotic Pressure in Biological Systems

• Normal Saline Solution: The fluid inside a blood cell has an osmotic pressure equivalent to that of a 0.9% (mass/volume) sodium chloride solution, also known as normal saline. This concentration is isotonic to the blood cells, meaning it is safe for intravenous injections as it does not cause the cells to shrink or swell.

• Hypertonic Solution: A solution with a higher concentration of solutes (e.g., more than 0.9% sodium chloride) is hypertonic relative to the inside of the blood cells. If blood cells are placed in such a solution, water will flow out of the cells, causing them to shrink (crenation).

• Hypotonic Solution: A solution with a lower concentration of solutes (e.g., less than 0.9% sodium chloride) is hypotonic relative to the inside of the blood cells. Placing blood cells in this solution will cause water to flow into the cells, leading them to swell and potentially burst (lysis).

Some Important Questions

Q.1. What are the properties arising due to varying concentrations of solute in a given solvent, irrespective of the nature of the solute concerning the solvent?

a) Colligative properties

b) Intensive properties

c) Extensive properties

d) Solute properties

Ans: a

Colligative properties are a set of four properties that depend only on the number of solute particles present in the solution and not on their chemical nature. The four properties are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

Q.2. Which law specifically governs the relative lowering of vapor pressures in solutions?

a) Van't Hoff law

b) Boyle's law

c) Raoult's law

d) Amagat's law

Ans: c

Raoult's law states p₁ = X₁ p₁^˚ for the solvent; the lowering Δp = p₁^˚ - p = p₁^˚(1 - X₁) = p₁^˚ X₂. Hence the relative lowering Δp/p₁^˚ equals the mole fraction of solute X₂.

Q.3. On addition of non-volatile potassium iodide in water at 298 K it is noticed that vapour pressure reduces from 23.8 mm Hg to 2.0 cm Hg. What is the mole fraction of solute in the solution?

a) 0.916

b) 0.160

c) 0.084

d) 0.092

Ans: b

Given p₀ (water) = 23.8 mm Hg; p (water after addition) = 2.0 cm Hg = 20.0 mm Hg

\( \Delta p = 23.8 - 20.0 = 3.8\ \text{mm Hg} \)

\( X_{\text{solute}} = \dfrac{\Delta p}{p^{\circ}} = \dfrac{3.8}{23.8} = 0.160 \)

Q.4. Which of the following is Raoult's law applicable to, to determine molar masses correctly?

a) Ionic solute in liquid

b) Non-ionic solute in dilute solution

c) Non-ionic solute in concentrated solution

d) Ionic solid in insoluble form in solvent

Ans: b

For accurate molar mass determination from colligative measurements, the solute must be non-volatile, non-ionic (non-dissociating) and the solution must be sufficiently dilute so that Raoult's law is obeyed. For ionic solutes, dissociation increases particle number and the Van't Hoff factor must be accounted for.

Q.5. Which of the following is a colligative property?

a) Relative lowering of fluid pressure

b) Decrease in boiling point

c) Decrease in freezing point

d) Change in volume after mixing

Ans: c

Depression in freezing point (decrease in freezing point) is a colligative property. The standard set consists of: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure.

Q.6. Mass of Urea (NH₂CONH₂) required to be dissolved in 1000 g of water in order to reduce the vapour pressure of water by 25% is _________ g. (Nearest integer)

Given: Molar mass of N, C, O and H are 14, 12, 16 and 1 g mol^-1 respectively

Ans: 1111

Solution.

Reduction of vapour pressure by 25% means vapour pressure of solution = 0.75 p^˚. Hence relative lowering = 0.25.

\( X_{\text{solute}} = 0.25 \)

If x is moles of urea and n is moles of water (1000 g ⇒ 1000/18 = 55.556 mol):

\( X_{\text{solute}} = \dfrac{x}{x + n} = 0.25 \)

\( x = \dfrac{0.25}{0.75} n = \dfrac{1}{3} n = \dfrac{1}{3} \times 55.556 = 18.519\ \text{mol} \)

Molar mass of urea = 14×2 + 12 + 16 + 1×4 = 60 g mol^-1

Mass required = moles × M = 18.519 × 60 = 1111 g