Colligative Properties: Why is osmotic pressure a better method for determining the molar mass of biomolecules like proteins

Chapter: Solutions – Colligative Properties

📘 Osmotic Pressure

Osmotic pressure (π) is the minimum external pressure that must be applied to a solution to stop the flow of solvent molecules through a semipermeable membrane.

π = CRT

For molar mass determination:

π = (w/MV)RT

Therefore,

M = wRT / (πV)

where w is the mass of solute, M is its molar mass, V is the volume of solution, R is the gas constant and T is the absolute temperature.

🔬 Why is it better for proteins and other biomolecules?

Osmotic pressure is particularly suitable for determining the molar mass of biomolecules such as proteins, enzymes and polymers because these substances generally have very high molar masses and may decompose when heated.

High Molar Mass
+
Heat Sensitive
+
Dilute Solutions
↓
Osmotic Pressure Method is Preferred

1. Suitable for Very High Molar Mass

For a given mass of a high-molar-mass biomolecule, the number of moles is very small. Other colligative properties such as elevation in boiling point or depression in freezing point may therefore be extremely small and difficult to measure accurately.

Osmotic pressure can be measured even for very dilute solutions, making it more sensitive for such substances.

2. No Heating is Required

Proteins and other biomolecules are often heat-sensitive. Methods involving boiling or freezing may cause decomposition, denaturation or structural changes.

Osmotic pressure can be determined at or near room temperature, so the biomolecule is less likely to undergo thermal decomposition.

3. Suitable for Dilute Solutions

Biomolecules can often be studied only in very dilute solutions. Osmotic pressure is directly proportional to the concentration:

π = CRT

Thus, even a dilute solution gives a measurable osmotic pressure.

4. Direct Relation with Molar Mass

For a non-electrolyte:

π = (w/MV)RT

Rearranging:

M = wRT / (πV)

Hence, the molar mass can be calculated directly from the measured osmotic pressure.

✅ Final Answer

Osmotic pressure is preferred for determining the molar mass of biomolecules such as proteins because it can be measured at room temperature using very dilute solutions.

Proteins have very high molar masses, so their other colligative effects such as elevation in boiling point and depression in freezing point are usually too small to measure accurately. Moreover, heating can denature or decompose proteins. Osmotic pressure avoids these problems and therefore gives a more reliable molar mass.

1. Osmotic pressure is a:

A) Colligative property

B) Chemical property

C) Nuclear property

D) Optical property

✅ Answer

A) Colligative property

---

2. The equation for osmotic pressure of a dilute solution is:

A) π = CRT

B) π = C/RT

C) π = RT/C

D) π = C + RT

✅ Answer

A) π = CRT

---

3. Osmotic pressure is especially useful for determining the molar mass of:

A) Small inorganic salts only

B) Proteins and polymers

C) Metals

D) Noble gases

✅ Answer

B) Proteins and polymers

---

4. Proteins are preferably studied by osmotic pressure because:

A) They have very low molar mass

B) They are often heat-sensitive

C) They are gases

D) They are always ionic

✅ Answer

B) They are often heat-sensitive

---

5. Which process may damage proteins?

A) Heating

B) Dilution

C) Measuring volume

D) Measuring pressure

✅ Answer

A) Heating

---

6. Osmotic pressure can be measured using:

A) Very concentrated solutions only

B) Dilute solutions

C) Solids only

D) Gases only

✅ Answer

B) Dilute solutions

---

7. The molar mass calculated using osmotic pressure is:

A) M = πV/wRT

B) M = wRT/πV

C) M = πRT/wV

D) M = wπ/RTV

✅ Answer

B) M = wRT/πV

---

8. Osmotic pressure is measured across a:

A) Metallic sheet

B) Semipermeable membrane

C) Glass plate

D) Filter paper only

✅ Answer

B) Semipermeable membrane

---

9. For a dilute solution, osmotic pressure is proportional to:

A) Temperature

B) Concentration

C) Both A and B

D) Neither A nor B

✅ Answer

C) Both A and B

---

10. Which property is generally very small for high-molar-mass proteins?

A) Osmotic pressure

B) Elevation in boiling point

C) Mass

D) Volume

✅ Answer

B) Elevation in boiling point

---

11. Another name for osmotic pressure equation π = CRT is:

A) Boyle equation

B) van't Hoff equation

C) Dalton equation

D) Graham equation

✅ Answer

B) van't Hoff equation

---

12. Osmotic pressure is measured at:

A) Only the boiling point

B) Only the freezing point

C) Convenient temperature, often near room temperature

D) Absolute zero

✅ Answer

C

---

13. Proteins generally have:

A) Very low molar masses

B) Very high molar masses

C) Zero molar mass

D) Fixed molar mass of 18 g mol⁻¹

✅ Answer

B) Very high molar masses

---

14. Which property does not require boiling the solution?

A) Osmotic pressure

B) Elevation in boiling point

C) Both always require boiling

D) None

✅ Answer

A) Osmotic pressure

---

15. The unit of osmotic pressure can be:

A) atm

B) mol kg⁻¹

C) mol L⁻¹

D) g mol⁻¹

✅ Answer

A) atm

---

16. Osmotic pressure depends mainly on the:

A) Number of solute particles

B) Colour of solute

C) Shape of container

D) Colour of solvent

✅ Answer

A) Number of solute particles

---

17. Which is an advantage of osmotic pressure for biomolecules?

A) It requires high temperature

B) It can be measured in dilute solutions

C) It requires boiling

D) It decomposes proteins

✅ Answer

B

---

18. The molar mass of a solute increases when, keeping w, T and V constant, osmotic pressure:

A) Increases

B) Decreases

C) Becomes zero only

D) Remains unrelated

✅ Answer

B) Decreases

---

19. Osmotic pressure is a property of:

A) Only pure solvents

B) Solutions

C) Solids only

D) Metals only

✅ Answer

B) Solutions

---

20. The best reason for using osmotic pressure for proteins is:

A) Proteins boil easily

B) Osmotic pressure can be measured without heating and in dilute solutions

C) Proteins have no mass

D) Proteins are volatile

✅ Answer

B

Q. Why is osmotic pressure preferred for determining the molar mass of proteins?

Answer: Proteins have very high molar masses and are often heat-sensitive. Osmotic pressure can be measured at relatively low temperature using very dilute solutions, whereas other colligative effects may be too small to measure accurately.

Q. Give three reasons why osmotic pressure is suitable for determining the molar mass of biomolecules.

1. Biomolecules have very high molar masses.
2. Osmotic pressure can be measured using dilute solutions.
3. No heating or boiling is required, reducing the possibility of denaturation or decomposition.

Q. Derive the expression used to determine the molar mass of a biomolecule from osmotic pressure.

For a dilute solution:

π = CRT

Since concentration:

C = n/V = w/MV

Therefore:

π = (w/MV)RT

Rearranging:

M = wRT / πV

Thus, the molar mass can be calculated from the mass, volume, temperature and measured osmotic pressure.

Q. Explain why osmotic pressure is a better method than elevation in boiling point or depression in freezing point for determining the molar mass of proteins.

Solution:

1. Proteins and other biomolecules generally possess very high molar masses.
2. For a given mass, their number of moles is very small.
3. Consequently, their elevation in boiling point and depression in freezing point are extremely small and difficult to measure accurately.
4. Osmotic pressure can be measured even in very dilute solutions.
5. The measurement can be performed without heating, so heat-sensitive proteins are less likely to denature or decompose.

Therefore, osmotic pressure provides a more convenient and reliable method for determining the molar mass of biomolecules.

Q. Explain in detail why osmotic pressure is preferred for determining the molar mass of biomolecules such as proteins.

1. High Molar Mass

Proteins, enzymes and many polymers have very high molar masses. Therefore, even when a measurable mass is taken, the number of moles present is small.

2. Very Small Other Colligative Effects

The elevation in boiling point and depression in freezing point depend on the molality of the solution. For high-molar-mass substances, the molality is often extremely small, giving very small temperature changes.

3. Osmotic Pressure is Measurable in Dilute Solutions

π = CRT

Even a dilute solution can produce a measurable osmotic pressure. This makes the method suitable for biomolecules that are commonly studied in dilute solutions.

4. No Heating Required

Many proteins are heat-sensitive and may undergo denaturation or decomposition at elevated temperatures. Osmotic pressure can be measured without boiling the solution.

5. Direct Calculation of Molar Mass

π = (w/MV)RT

M = wRT/(πV)

Thus, the molar mass can be calculated directly from the experimentally measured osmotic pressure.

Conclusion:

Osmotic pressure is preferred because it is sensitive for dilute solutions, can be measured without heating, and is therefore especially suitable for high-molar-mass, heat-sensitive biomolecules such as proteins.

🎯 Quick Revision

Osmotic pressure: π = CRT
Molar mass: M = wRT/(πV)
Proteins: High molar mass + heat-sensitive
Solution: Very dilute solution can be used
Temperature: No boiling/heating required
Result: More reliable molar-mass determination

Azeotropes: Explain why certain liquid mixtures form minimum boiling azeotropes

📚 Chapter: Solutions – Azeotropes

📘 Definition of Azeotrope

An azeotrope is a liquid mixture that boils at a constant temperature and whose liquid and vapour phases have the same composition.

Liquid composition = Vapour composition
↓
Constant-boiling mixture

Because the vapour has the same composition as the liquid, an azeotrope cannot be separated into its components by ordinary fractional distillation.

🔥 What is a Minimum-Boiling Azeotrope?

A minimum-boiling azeotrope is a mixture that boils at a temperature lower than the boiling points of either of its pure components.

Boiling point of azeotrope
< Boiling point of A
< Boiling point of B

Such azeotropes are formed when the solution shows a positive deviation from Raoult's law.

🔬 Why does Positive Deviation occur?

In a minimum-boiling azeotrope, the attractive forces between unlike molecules A–B are weaker than the attractive forces between like molecules A–A and B–B.

A–B attraction
<
A–A and B–B attractions

Therefore, molecules escape more easily from the liquid phase into the vapour phase.

This produces a higher vapour pressure than predicted by Raoult's law.

Weaker A–B interactions
↓
Greater escape tendency
↓
Higher vapour pressure
↓
Lower boiling temperature
↓
Minimum-Boiling Azeotrope

📌 Important Example

A common example is the mixture of ethanol and water.

The ethanol–water system forms a minimum-boiling azeotrope at approximately 95.6% ethanol by mass under atmospheric pressure. Its boiling temperature is about 78.2°C, slightly below the boiling point of pure ethanol.

Therefore, ordinary fractional distillation cannot produce completely anhydrous ethanol from this azeotropic mixture.

✅ Final Answer

Minimum-boiling azeotropes are formed due to a positive deviation from Raoult's law.

The unlike molecular interactions (A–B) are weaker than the like interactions (A–A and B–B), causing the vapour pressure to increase. Consequently, the mixture boils at a temperature lower than either pure component.

1. An azeotrope is a mixture in which:

A) Only liquid exists

B) Liquid and vapour compositions are the same

C) Only vapour exists

D) Components are completely immiscible

✅ Answer

B) Liquid and vapour compositions are the same

---

2. A minimum-boiling azeotrope shows:

A) Positive deviation from Raoult's law

B) Negative deviation from Raoult's law

C) No deviation

D) Ideal behaviour

✅ Answer

A

---

3. In a minimum-boiling azeotrope, vapour pressure is:

A) Lower than expected

B) Higher than expected

C) Zero

D) Unchanged

✅ Answer

B

---

4. The boiling point of a minimum-boiling azeotrope is:

A) Higher than both components

B) Lower than both components

C) Equal to both components

D) Always 100°C

✅ Answer

B

---

5. Which type of intermolecular interaction produces positive deviation?

A) Stronger A–B interactions

B) Weaker A–B interactions

C) No molecular interaction

D) Ionic bonding only

✅ Answer

B

---

6. A minimum-boiling azeotrope is also called:

A) Maximum-boiling azeotrope

B) Constant-boiling mixture

C) Pure solvent

D) Saturated solution

✅ Answer

B

---

7. Which mixture is a well-known minimum-boiling azeotrope?

A) Ethanol–water

B) NaCl–water

C) Sugar–water

D) Cu–Zn

✅ Answer

A) Ethanol–water

---

8. Positive deviation causes the total vapour pressure to:

A) Increase

B) Decrease

C) Become zero

D) Remain constant

✅ Answer

A

---

9. Minimum-boiling azeotropes cannot be separated by:

A) Ordinary fractional distillation

B) Chemical reaction

C) Special separation methods

D) None

✅ Answer

A

---

10. At the azeotropic composition:

A) Liquid and vapour compositions are equal

B) Only liquid composition exists

C) Only vapour composition exists

D) No equilibrium exists

✅ Answer

A

---

11. The A–B interactions in a positive deviation are generally:

A) Stronger than A–A and B–B

B) Weaker than A–A and B–B

C) Equal to zero

D) Always ionic

✅ Answer

B

---

12. Higher vapour pressure generally means:

A) Higher boiling point

B) Lower boiling point

C) No boiling

D) Higher melting point

✅ Answer

B

---

13. Minimum-boiling azeotrope occurs at a composition where:

A) Liquid and vapour compositions differ greatly

B) Liquid and vapour compositions are identical

C) No vapour is present

D) No liquid is present

✅ Answer

B

---

14. Which law describes ideal liquid-solution behaviour?

A) Boyle's law

B) Raoult's law

C) Charles' law

D) Graham's law

✅ Answer

B

---

15. A minimum-boiling azeotrope has a boiling point that is:

A) Minimum at the azeotropic composition

B) Maximum at the azeotropic composition

C) Always zero

D) Independent of composition

✅ Answer

A

---

16. Which deviation is associated with weaker intermolecular forces?

A) Positive deviation

B) Negative deviation

C) No deviation

D) Ideal behaviour

✅ Answer

A

---

17. A minimum-boiling azeotrope is generally:

A) Less volatile than both components

B) More volatile than either component at the azeotropic composition

C) Non-volatile

D) Solid

✅ Answer

B

---

18. Ethanol–water forms an azeotrope because of:

A) Ideal behaviour over all compositions

B) Deviation from Raoult's law

C) Formation of a solid

D) Complete ionisation

✅ Answer

B

---

19. The major consequence of azeotrope formation is:

A) Easy separation by ordinary distillation

B) Difficulty in separation by ordinary distillation

C) Complete precipitation

D) Formation of an ionic solid

✅ Answer

B

---

20. Which statement is correct?

A) Minimum-boiling azeotropes show negative deviation

B) Minimum-boiling azeotropes show positive deviation

C) Minimum-boiling azeotropes obey Raoult's law exactly

D) Minimum-boiling azeotropes have zero vapour pressure

✅ Answer

B) Minimum-boiling azeotropes show positive deviation

Q. Why do certain liquid mixtures form minimum-boiling azeotropes?

Solution: Minimum-boiling azeotropes are formed due to positive deviation from Raoult's law. The attractive forces between unlike molecules are weaker than those between like molecules, causing higher vapour pressure and consequently a lower boiling point.

Q. Explain the formation of a minimum-boiling azeotrope.

Solution:

1. The mixture shows positive deviation from Raoult's law.
2. A–B intermolecular attractions are weaker than A–A and B–B attractions.
3. The vapour pressure becomes higher and the boiling point becomes lower than that of either pure component.

Hence, a minimum-boiling azeotrope is formed.

Q. Explain the relationship between positive deviation and minimum-boiling azeotrope.

Positive Deviation
↓
Weaker A–B attraction
↓
Higher Vapour Pressure
↓
Lower Boiling Point
↓
Minimum-Boiling Azeotrope

Thus, a minimum-boiling azeotrope occurs when the total vapour pressure shows a maximum and the boiling temperature shows a minimum at a particular composition.

Q. Explain minimum-boiling azeotropes with an example.

Solution: A minimum-boiling azeotrope is a constant-boiling mixture that boils at a temperature lower than the boiling points of both pure components.

It is formed when the solution shows positive deviation from Raoult's law. The A–B attractions are weaker than A–A and B–B attractions. Consequently, molecules escape more readily from the liquid phase, increasing the vapour pressure.

Since boiling occurs when vapour pressure becomes equal to external pressure, the higher vapour pressure causes the solution to boil at a lower temperature.

Example: Ethanol–water forms a minimum-boiling azeotrope under atmospheric pressure.

Weaker A–B interactions → Positive deviation → Higher vapour pressure → Lower boiling point

Q. Explain in detail why certain liquid mixtures form minimum-boiling azeotropes.

1. Azeotrope

An azeotrope is a constant-boiling mixture in which the composition of the vapour phase is the same as that of the liquid phase at the azeotropic composition.

2. Positive Deviation from Raoult's Law

Minimum-boiling azeotropes are associated with positive deviation from Raoult's law. The observed vapour pressure is greater than the vapour pressure predicted for an ideal solution.

3. Weak A–B Interactions

The intermolecular attractions between unlike molecules, A–B, are weaker than the attractions between like molecules, A–A and B–B.

4. Increase in Vapour Pressure

Because the A–B attractions are weaker, molecules can escape from the liquid phase more easily. Therefore, the vapour pressure becomes relatively high.

5. Decrease in Boiling Point

A liquid boils when its vapour pressure becomes equal to the external pressure. Since the vapour pressure is higher, the mixture reaches the boiling condition at a lower temperature.

Weak A–B attraction
↓
Positive deviation
↓
Higher vapour pressure
↓
Lower boiling temperature
↓
Minimum-boiling azeotrope

6. Example

The ethanol–water system is a well-known example of a minimum-boiling azeotrope under atmospheric pressure.

Final Answer:

Certain liquid mixtures form minimum-boiling azeotropes because of positive deviation from Raoult's law, caused by weaker unlike-molecule interactions. This gives a higher vapour pressure and hence a lower boiling temperature at the azeotropic composition.

🎯 Quick Revision

Minimum-boiling azeotrope → Positive deviation
A–B interaction → Weaker
Vapour pressure → Higher
Boiling point → Lower
Example → Ethanol–water
Separation by ordinary fractional distillation → Not possible at azeotropic composition

Solutions: Calculate the molality of a solution where the vapour pressure is lowered by a non-volatile solute

📚 Chapter: Solutions – Colligative Properties

📘 Concept

When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solution becomes lower than that of the pure solvent. This is called lowering of vapour pressure.

Relative Lowering of Vapour Pressure

ΔP / P° = X_solute

For a dilute solution:

X_solute ≈ m × M_solvent / 1000

Therefore, the molality is:

m = (ΔP / P°) × (1000 / M_solvent)

where:

• ΔP = P° − P
• P° = vapour pressure of pure solvent
• P = vapour pressure of solution
• M_solvent = molar mass of solvent in g mol⁻¹
• m = molality in mol kg⁻¹

Question:

The vapour pressure of pure water at a certain temperature is 24.0 mm Hg, while the vapour pressure of a solution containing a non-volatile solute is 23.5 mm Hg. Calculate the molality of the solution.

Step 1: Calculate lowering of vapour pressure

ΔP = P° − P
= 24.0 − 23.5
= 0.5 mm Hg

Step 2: Calculate relative lowering

ΔP / P° = 0.5 / 24.0
= 0.02083

Step 3: Use the dilute-solution relation

For water, molar mass: M_H₂O = 18 g mol⁻¹

m = (0.02083 × 1000) / 18
= 1.157 mol kg⁻¹

Answer:

Molality ≈ 1.16 mol kg⁻¹

1. Addition of a non-volatile solute to a solvent causes vapour pressure to:

A) Increase

B) Decrease

C) Remain unchanged

D) Become zero always

✅ Answer

B) Decrease

---

2. Lowering of vapour pressure is a:

A) Colligative property

B) Chemical property

C) Nuclear property

D) Optical property

✅ Answer

A) Colligative property

---

3. For a non-volatile solute, relative lowering of vapour pressure is equal to:

A) Mole fraction of solvent

B) Mole fraction of solute

C) Molality of solvent

D) Molarity of solvent

✅ Answer

B) Mole fraction of solute

---

4. Molality is expressed as:

A) mol L⁻¹

B) mol kg⁻¹

C) g L⁻¹

D) g mol⁻¹

✅ Answer

B) mol kg⁻¹

---

5. The vapour pressure of pure solvent is represented by:

A) P

B) P°

C) ΔP

D) X

✅ Answer

B) P°

---

6. Lowering of vapour pressure is:

A) P − P°

B) P° − P

C) P° + P

D) P/P°

✅ Answer

B) P° − P

---

7. If P° = 30 mm Hg and P = 29 mm Hg, ΔP is:

A) 1 mm Hg

B) 29 mm Hg

C) 30 mm Hg

D) 59 mm Hg

✅ Answer

A) 1 mm Hg

---

8. Raoult's law for a non-volatile solute is:

A) P = P°X_solute

B) P = P°X_solvent

C) P = P° + X

D) P = X/P°

✅ Answer

B) P = P°X_solvent

---

9. If the mole fraction of solute increases, vapour pressure generally:

A) Increases

B) Decreases

C) Remains constant

D) Becomes infinite

✅ Answer

B) Decreases

---

10. Which quantity is independent of temperature for a given mass of solvent?

A) Molarity

B) Molality

C) Vapour pressure

D) Volume

✅ Answer

B) Molality

---

11. For water, the molar mass is:

A) 16 g mol⁻¹

B) 17 g mol⁻¹

C) 18 g mol⁻¹

D) 20 g mol⁻¹

✅ Answer

C) 18 g mol⁻¹

---

12. Molality is calculated using the mass of:

A) Solution

B) Solute

C) Solvent

D) Both solute and solvent equally

✅ Answer

C) Solvent

---

13. The formula m = (ΔP/P°)(1000/M) is used for:

A) Dilution only

B) Finding molality from vapour-pressure lowering

C) Finding density

D) Finding pH

✅ Answer

B

---

14. A non-volatile solute:

A) Contributes significantly to vapour pressure

B) Does not contribute appreciably to vapour pressure

C) Always evaporates first

D) Increases solvent vapour pressure

✅ Answer

B

---

15. If the relative lowering of vapour pressure is 0.01, the mole fraction of solute for a dilute solution is approximately:

A) 0.1

B) 0.01

C) 1.0

D) 10

✅ Answer

B) 0.01

---

16. The unit of vapour pressure is commonly:

A) mm Hg

B) mol kg⁻¹

C) mol L⁻¹

D) g mol⁻¹

✅ Answer

A) mm Hg

---

17. Which law describes vapour pressure of an ideal solution?

A) Boyle's law

B) Charles' law

C) Raoult's law

D) Graham's law

✅ Answer

C) Raoult's law

---

18. If P decreases while P° remains constant, ΔP:

A) Decreases

B) Increases

C) Remains zero

D) Cannot be calculated

✅ Answer

B) Increases

---

19. A solution with greater concentration of a non-volatile solute generally has:

A) Greater vapour pressure

B) Lower vapour pressure

C) No vapour pressure

D) Same vapour pressure as pure solvent

✅ Answer

B) Lower vapour pressure

---

20. Molality depends on:

A) Volume of solution

B) Mass of solvent

C) Temperature-dependent volume

D) Vapour pressure only

✅ Answer

B) Mass of solvent

Q. How can molality be calculated from lowering of vapour pressure?

Solution: For a dilute solution containing a non-volatile solute:

m = (ΔP / P°) × (1000 / M_solvent)

where ΔP = P° − P.

Q. A solution has P° = 25 mm Hg and P = 24.5 mm Hg. Calculate its relative lowering of vapour pressure.

Solution:

ΔP = 25 − 24.5 = 0.5 mm Hg

ΔP/P° = 0.5/25 = 0.020

Therefore, the relative lowering of vapour pressure is 0.020.

Q. Derive the expression for calculating molality from relative lowering of vapour pressure.

For a dilute solution containing a non-volatile solute:

ΔP/P° = X_solute

For a dilute solution:

X_solute ≈ n_solute/n_solvent

Since

m = n_solute / mass of solvent (kg)

and for 1 kg solvent:

n_solvent = 1000/M_solvent

Therefore:

m = (ΔP/P°) × (1000/M_solvent)

Q. Explain how the molality of a solution can be determined from its lowering of vapour pressure.

Solution: When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solvent decreases. According to Raoult's law:

P = P°X_solvent

Hence:

ΔP/P° = X_solute

For a dilute solution, the mole fraction of solute can be related to molality:

X_solute ≈ mM_solvent/1000

Therefore:

m = (ΔP/P°) × (1000/M_solvent)

This equation allows the molality to be calculated when the vapour pressure of the pure solvent, vapour pressure of the solution and molar mass of the solvent are known.

Q. Derive the relation between lowering of vapour pressure and molality for a solution containing a non-volatile solute.

Step 1: Raoult's Law

For a solution containing a non-volatile solute, the vapour pressure of the solution is:

P = P°X_solvent

Step 2: Lowering of Vapour Pressure

ΔP = P° − P

Therefore:

ΔP/P° = X_solute

Step 3: Relate Mole Fraction to Molality

For a dilute solution:

X_solute ≈ n_solute/n_solvent

If the mass of solvent is expressed in kilograms, then:

m = n_solute/kg of solvent

For 1 kg of solvent:

n_solvent = 1000/M_solvent

Thus:

X_solute ≈ mM_solvent/1000

Step 4: Final Relation

ΔP/P° = mM_solvent/1000

Therefore:

m = (ΔP/P°) × (1000/M_solvent)

Conclusion: The molality of a dilute solution containing a non-volatile solute can be determined from its relative lowering of vapour pressure.

🎯 Quick Revision

Lowering: ΔP = P° − P
Raoult's law: ΔP/P° = X_solute
Molality: m = moles of solute / kg of solvent
Useful relation: m = (ΔP/P°)(1000/M_solvent)
Unit: mol kg⁻¹