Why does the addition of a non-volatile solute lower the vapor pressure of a solvent?

Solute particles occupy a fraction of the liquid surface area, reducing the effective number of solvent molecules exposed to the surface per unit area. This directly decreases the rate of solvent vaporization. Since vapor pressure is the dynamic equilibrium pressure between evaporation and condensation rates, lowering the evaporation rate reduces equilibrium vapor pressure $P < P^\circ$.

Why is osmotic pressure measurement preferred over freezing point depression for determining the molar mass of biomolecules like proteins?

Macromolecules and proteins have large molar masses ($M > 10^4\text{ g mol}^{-1}$), yielding extremely small molal concentrations in solution. This results in immeasurably tiny freezing point depressions ($\Delta T_f < 10^{-3}\text{ K}$). In contrast, osmotic pressure $\pi = C R T$ generates a significant, easily measurable fluid head height in millimeters even at room temperature $T = 298\text{ K}$ for very low molar concentrations.

How do I identify whether a binary mixture will form a maximum or minimum boiling azeotrope?

Examine the relative strength of intermolecular forces. If $A-B$ attractive forces are stronger than $A-A$ and $B-B$ forces, the solution shows negative deviation from Raoult's law ($P_{\text{total}}$ is lower than predicted). A lower vapor pressure means a higher boiling point is required to vaporize the mixture, forming a maximum boiling azeotrope at a specific composition.

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Q: Why is osmotic pressure measurement preferred over freezing point depression for determining the molar mass of biomolecules like proteins?

Macromolecules and proteins have large molar masses ($M > 10^4\text{ g mol}^{-1}$), yielding extremely small molal concentrations in solution. This results in immeasurably tiny freezing point depressions ($\Delta T_f < 10^{-3}\text{ K}$). In contrast, osmotic pressure $\pi = C R T$ generates a significant, easily measurable fluid head height in millimeters even at room temperature $T = 298\text{ K}$ for very low molar concentrations.

Q: How do I identify whether a binary mixture will form a maximum or minimum boiling azeotrope?

Examine the relative strength of intermolecular forces. If $A-B$ attractive forces are stronger than $A-A$ and $B-B$ forces, the solution shows negative deviation from Raoult's law ($P_{\text{total}}$ is lower than predicted). A lower vapor pressure means a higher boiling point is required to vaporize the mixture, forming a maximum boiling azeotrope at a specific composition.

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Question #1NEET 2024

At $300\text{ K}$ , the osmotic pressure of a $0.1\text{ M}$ aqueous solution of a weak monobasic acid $\text{HA}$ is measured as $2.952\text{ atm}$ . What is the degree of dissociation $\alpha$ of the acid? (Take $R = 0.0821\text{ L atm mol}^{-1}\text{K}^{-1}$ )

Solution Locked

Step-by-Step Solution & Concept

Step 1

Write the osmotic pressure formula including the van 't Hoff factor: $\pi = i \cdot C \cdot R \cdot T$ .

Step 2

Substitute given values to find $i$ : $2.952 = i \times 0.1 \times 0.0821 \times 300 \implies 2.952 = i \times 2.463 \implies i = \frac{2.952}{2.463} = 1.20$ .

Step 3

For monobasic acid dissociation $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ , total ions per formula unit $n = 2$ . Using $i = 1 + (n - 1)\alpha \implies 1.20 = 1 + (2 - 1)\alpha \implies \alpha = 0.20$ .

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Question #2JEE Main 2023

Two liquids $A$ and $B$ form an ideal solution. At $300\text{ K}$ , the vapor pressure of pure liquid $A$ is $100\text{ mm Hg}$ and that of pure liquid $B$ is $200\text{ mm Hg}$ . If a liquid mixture contains $2\text{ moles}$ of $A$ and $3\text{ moles}$ of $B$ , determine the mole fraction of component $A$ in the vapor phase ( $Y_A$ ) at equilibrium.

Solution Locked

Step-by-Step Solution & Concept

Step 1

Compute mole fractions in the liquid phase: $X_A = \frac{2}{2 + 3} = 0.40$ and $X_B = \frac{3}{2 + 3} = 0.60$ .

Step 2

Calculate partial pressures using Raoult's Law: $P_A = P_A^\circ X_A = 100 \times 0.40 = 40\text{ mm Hg}$ and $P_B = P_B^\circ X_B = 200 \times 0.60 = 120\text{ mm Hg}$ . Total vapor pressure $P_{\text{total}} = P_A + P_B = 160\text{ mm Hg}$ .

Step 3

Apply Dalton's law of partial pressures to find the vapor phase mole fraction: $Y_A = \frac{P_A}{P_{\text{total}}} = \frac{40}{160} = 0.25$ .

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Raoult's Law for Relative Lowering of Vapor Pressure

Elevation in Boiling Point

Depression in Freezing Point

Osmotic Pressure Equation

Van 't Hoff Factor for Dissociation/Association

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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