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

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2. Lowering of vapour pressure is a:

A) Colligative property

B) Chemical property

C) Nuclear property

D) Optical property

✅ Answer

A) Colligative property

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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

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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

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

A) P

B) P°

C) ΔP

D) X

✅ Answer

B) P°

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6. Lowering of vapour pressure is:

A) P − P°

B) P° − P

C) P° + P

D) P/P°

✅ Answer

B) P° − P

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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

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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

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

A) Increases

B) Decreases

C) Remains constant

D) Becomes infinite

✅ Answer

B) Decreases

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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

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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

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12. Molality is calculated using the mass of:

A) Solution

B) Solute

C) Solvent

D) Both solute and solvent equally

✅ Answer

C) Solvent

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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

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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

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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

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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

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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

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

A) Decreases

B) Increases

C) Remains zero

D) Cannot be calculated

✅ Answer

B) Increases

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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

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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