We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001

NCERT Exemplar · Q20

Q. We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van't Hoff factor for these solutions will be in the order______.

(i)

i_A < i_B < i_C

(ii)

i_A > i_B > i_C

(iii)

i_A = i_B = i_C

(iv)

i_A < i_B > i_C

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Concept understanding — VanT Hoff Factor Association

The Intuition: What Happens When Particles Stick Together?

Imagine you're counting people in a room. You see 100 chairs, each with one person. That's 100 individuals. Now imagine those same 100 people decide to pair up — every two people hold hands and become a "couple." Suddenly, the number of independent moving units in the room drops from 100 to 50.

That's exactly what association does in a solution. When solute particles (molecules or ions) associate, they clump together into larger clusters. The number of independent particles floating around decreases. And since colligative properties (freezing point depression, boiling point elevation, osmotic pressure) depend only on the number of particles — not their identity — the observed effect becomes smaller than expected.

Note

Association is the opposite of dissociation. In dissociation, one particle breaks into many (e.g., NaCl → Na⁺ + Cl⁻). In association, many particles combine into one (e.g., two acetic acid molecules dimerise).

The Van't Hoff Factor: The Correction Number

The Van't Hoff factor, denoted by $i$ , is defined as:

i = \frac{\text{Actual number of particles in solution after association}}{\text{Number of particles if no association occurred}}

For a non-electrolyte that does not associate or dissociate, $i = 1$ .

For association, $i < 1$ — because the actual particle count is less than what you started with.

A Concrete Example: Acetic Acid in Benzene

Acetic acid ( $\text{CH}_3\text{COOH}$ ) in benzene forms dimers — two molecules stick together via hydrogen bonding:

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

Suppose you dissolve 100 molecules of acetic acid. If no association occurred, you'd have 100 particles. But if all of them dimerise, you get only 50 dimers. So:

i = \frac{50}{100} = 0.5

In reality, association is never 100% complete — it's an equilibrium. So $i$ lies between 0.5 and 1.

Watch out

A common mistake: thinking $i$ can be negative. It cannot. For association, $0 < i < 1$ . For dissociation, $i > 1$ . For no change, $i = 1$ .

The General Formula for Association

Let’s say $n$ molecules of a solute associate to form one associated particle:

nA \rightleftharpoons A_n

Let $\alpha$ be the degree of association — the fraction of original molecules that have associated.

Initially: 1 mole of A (i.e., $N$ molecules)
Moles that associate: $\alpha$
Moles that remain as single A: $1 - \alpha$
Moles of associated particles formed: $\frac{\alpha}{n}$ (because $n$ molecules make 1 associated unit)

Total moles after association:

(1 - \alpha) + \frac{\alpha}{n}

The Van't Hoff factor is:

i = \frac{\text{Total moles after association}}{\text{Initial moles}} = \frac{(1 - \alpha) + \frac{\alpha}{n}}{1} = 1 - \alpha + \frac{\alpha}{n}

Simplify:

i = 1 - \alpha\left(1 - \frac{1}{n}\right)

For the common case of dimerisation ( $n = 2$ ):

i = 1 - \alpha\left(1 - \frac{1}{2}\right) = 1 - \frac{\alpha}{2}

So if $\alpha = 0.6$ (60% association), then $i = 1 - 0.3 = 0.7$ .

How Association Affects Colligative Properties

All colligative properties are multiplied by $i$ :

Property	Formula without association	Formula with association
Relative lowering of vapour pressure	$\frac{p^\circ - p}{p^\circ} = x_B$	$\frac{p^\circ - p}{p^\circ} = i \cdot x_B$
Elevation in boiling point	$\Delta T_b = K_b \cdot m$	$\Delta T_b = i \cdot K_b \cdot m$

Why this formula?

Van't Hoff Factor for Association: Why the Formula Holds

The Van't Hoff factor ( $i$ ) for association describes how solute particles combine in solution, reducing the effective number of particles. Let's build the reasoning step-by-step.

1. The Core Idea: What Changes?

When a solute associates (e.g., two acetic acid molecules dimerize in benzene), the number of particles in solution decreases. The Van't Hoff factor is defined as:

i = \frac{\text{Actual number of particles in solution}}{\text{Number of particles if no association}}

For association, $i < 1$ .

2. Setting Up the Association Process

Consider a solute that associates to form $n$ molecules per aggregate (e.g., $n=2$ for dimerization). Let:

Initial moles of solute = $1$ mole (for simplicity)
Degree of association = $\alpha$ (fraction of solute that associates)

What happens to the particles?

Moles that associate = $\alpha$ (these combine into aggregates)
Moles that remain free = $1 - \alpha$

Each associated group of $n$ molecules becomes 1 aggregate particle. So:

Number of aggregates formed = $\frac{\alpha}{n}$
Number of free molecules = $1 - \alpha$

3. Total Particles After Association

Total moles of particles in solution:

\text{Total} = \underbrace{(1 - \alpha)}_{\text{free}} + \underbrace{\frac{\alpha}{n}}_{\text{aggregates}}

If no association ( $\alpha = 0$ ), total = $1$ mole of particles.

4. The Van't Hoff Factor Formula

By definition:

i = \frac{\text{Total particles after association}}{\text{Total particles if no association}} = \frac{(1 - \alpha) + \frac{\alpha}{n}}{1}

Thus:

\boxed{i = 1 - \alpha + \frac{\alpha}{n}}

5. Why This Makes Physical Sense

If $\alpha = 0$ (no association): $i = 1$ — particles behave independently.
If $\alpha = 1$ (complete association): $i = \frac{1}{n}$ — all molecules form $n$ -mers, so particle count drops by factor $n$ . …

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The van’t Hoff factor $i$ for NaCl increases as the solution becomes more dilute because fewer ion pairs form. So the order is $i_A < i_B < i_C$ , which corresponds to option (i).

Why the van’t Hoff factor depends on concentration

NaCl is a strong electrolyte — in theory it dissociates completely into Na⁺ and Cl⁻, giving $i = 2$ . But in reality, at higher concentrations, some ions come close enough to feel electrostatic attraction and temporarily pair up (ion pairing). This reduces the effective number of particles. The effect is strongest when the solution is concentrated, and it weakens as we dilute.

So the van’t Hoff factor is not a constant for a given solute — it approaches the theoretical value only in the limit of infinite dilution. For NaCl, $i$ is always slightly less than 2, and it gets closer to 2 as the concentration drops.

Watch out

A common mistake is to assume $i = 2$ for all concentrations of a strong electrolyte. That’s only true at infinite dilution. In real solutions, ion pairing lowers $i$ , especially at higher molarities.

Step-by-step reasoning

Recall the definition The van’t Hoff factor $i$ is the ratio of the actual number of particles in solution to the number of formula units dissolved. For NaCl, if dissociation were complete:

i = \frac{\text{moles of particles}}{\text{moles of NaCl}} = 2

Identify the real behaviour

At finite concentrations, some Na⁺ and Cl⁻ ions associate transiently into ion pairs (Na⁺Cl⁻). These pairs count as one particle, not two. So the actual particle count is less than $2n$ , meaning $i < 2$ .
Connect concentration to ion pairing …

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Common Mistakes Students Make on van't Hoff Factor & Concentration

Mistake 1: Assuming $i$ is constant for all concentrations of the same solute

Why it happens: Students memorize that for NaCl, $i = 2$ (complete dissociation into Na⁺ and Cl⁻) and apply it blindly.

The truth: The van't Hoff factor $i$ depends on degree of dissociation, which changes with concentration. At infinite dilution, dissociation is complete ( $i \to 2$ ). At higher concentrations, ion pairing reduces $i$ .

How to avoid: Always ask: "Is this at infinite dilution or a real concentration?" For real solutions, $i$ decreases as concentration increases.

Mistake 2: Thinking higher concentration means higher $i$

Why it happens: Students confuse "more ions present" with "higher $i$ ". More concentrated solutions do have more ions per litre, but $i$ is a ratio:

i = \frac{\text{actual number of particles}}{\text{number of formula units dissolved}}

The truth: At higher concentration, ions are closer together and more likely to recombine (ion pairing), so the degree of dissociation decreases, making $i$ smaller.

How to avoid: Remember: Dilution favours dissociation. As concentration decreases, $i$ approaches the theoretical maximum.

Mistake 3: Picking option (iii) — $i_A = i_B = i_C$

Why it happens: Students think "same solute, same $i$ ".

The truth: $i$ is concentration-dependent. For NaCl:

At 0.1 M: significant ion pairing → $i \approx 1.87$
At 0.01 M: less pairing → $i \approx 1.94$
At 0.001 M: nearly complete dissociation → $i \approx 1.99$

How to avoid: Visualise the trend: more dilute → more dissociation → higher $i$ .

Mistake 4: Picking option (ii) — $i_A > i_B > i_C$

Why it happens: Students think "more concentrated means more ions, so higher $i$ ".

The truth: This reverses the actual trend. The correct order is:

\boxed{i_A < i_B < i_C}

which corresponds to option (i). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 21 on this concept.

TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
$Q. Identify the edge lengths (a, b, c) in the unit cell shown below [FIGURE] (A) i = a; j = b; k = c (B) i = a; j = c; k = b (C) i = c; j = b; k = a (D) i = b; j = c; k = a$
›Reveal solutionSolution
The key is to match the crystallographic axes (a, b, c) to the edges of the unit cell by the standard convention: a along x, b along y, c along z. In the figure, i is the horizontal edge (x-direction), j is the vertical edge (y-direction), and k is the depth edge (z-direction). Thus i = a, j = b, k = c, so the correct option is (A).

Concept and Intuition

In crystallography, the unit cell is defined by three edge vectors: a, b, and c. By convention, these are aligned with the Cartesian axes: a points along the x-direction (usually horizontal in a 2D drawing), b along the y-direction (vertical), and c along the z-direction (depth or out-of-page). The figure labels the edges as i, j, and k. To answer, we simply identify which physical direction each label corresponds to and match it to the standard axis assignment.

Step-by-step reasoning

Identify the axes in the figure

The unit cell is drawn as a rectangular box. The edge labeled i runs horizontally (left to right). The edge labeled j runs vertically (up and down). The edge labeled k runs into the page (depth), shown as a receding line.

Recall the crystallographic convention

The standard assignment is:

a = edge along the x-axis (horizontal)

b = edge along the y-axis (vertical) …
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TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
$858\ \mathrm{S\ cm^2\ mol^{-1}}$
›Reveal solutionSolution
Kohlrausch’s law of independent migration of ions says the molar conductivity of a salt at infinite dilution is the sum of the molar conductivities of its constituent ions, each multiplied by its stoichiometric coefficient. For Al₂(SO₄)₃, this gives λ°(Al³⁺) = 189 S cm² mol⁻¹, so the correct option is (C).

Concept & Intuition

At infinite dilution, ions move independently — they don’t interact. So the total molar conductivity of a salt is simply the sum of the contributions from each ion, weighted by how many of that ion appear in the formula unit. This is Kohlrausch’s law.

For aluminum sulfate, the formula is Al₂(SO₄)₃. That means 2 Al³⁺ ions and 3 SO₄²⁻ ions per formula unit. If we know the total molar conductivity of the salt and the molar conductivity of the sulfate ion, we can solve for the aluminum ion’s molar conductivity.

Step-by-step solution

Write Kohlrausch’s law for Al₂(SO₄)₃ The molar conductivity at infinite dilution is:

$\Lambda^\circ_{\mathrm{Al_2(SO_4)_3}} = 2\,\lambda^\circ_{\mathrm{Al^{3+}}} + 3\,\lambda^\circ_{\mathrm{SO_4^{2-}}}$
This is because each mole of salt releases 2 moles of Al³⁺ and 3 moles of SO₄²⁻.

Insert the known values We are given:

$\Lambda^\circ_{\mathrm{Al_2(SO_4)_3}} = 858\ \mathrm{S\ cm^2\ mol^{-1}}, \quad \lambda^\circ_{\mathrm{SO_4^{2-}}} = 160\ \mathrm{S\ cm^2\ mol^{-1}}$
So:
$858 = 2\,\lambda^\circ_{\mathrm{Al^{3+}}} + 3 \times 160$

Simplify and solve First, $3 \times 160 = 480$ . Then:
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$^-$
›Reveal solutionSolution
Fluoride is added to drinking water to prevent tooth decay, but too much can cause dental and skeletal problems. The maximum safe limit for fluoride in drinking water is $1.0 \text{ mg L}^{-1}$ to $2.0 \text{ mg L}^{-1}$ .

The question asks for the maximum limit of soluble fluoride salt that can be added to water to make it safe for drinking. This involves understanding the dual nature of fluoride: it is beneficial at low concentrations but harmful at higher concentrations.

Concept and Intuition

Fluoride ions ( $F^-$ ) are naturally present in some water sources and are also intentionally added to public water supplies in many regions. The primary reason for adding fluoride is its ability to strengthen tooth enamel, making teeth more resistant to decay caused by acids produced by bacteria. This is a crucial public health measure for preventing dental cavities.

However, like many essential trace elements, fluoride has a narrow optimal concentration range. If the concentration of fluoride in drinking water is too low, it doesn't provide sufficient protection against tooth decay. Conversely, if the concentration is too high, it can lead to adverse health effects, primarily affecting teeth and bones. Therefore, there is a carefully determined maximum limit to ensure the water remains safe and beneficial.

Beneficial Effects of Fluoride:

At low concentrations, typically around $1.0 \text{ mg L}^{-1}$ (or $1.0 \text{ ppm}$ ), fluoride ions integrate into the crystal structure of tooth enamel, forming fluoroapatite. Fluoroapatite is more stable and resistant to acid dissolution than the naturally occurring hydroxyapatite, thereby protecting teeth from cavities.

Optimal Concentration for Dental Health:

The optimal concentration of fluoride in drinking water for preventing tooth decay is generally considered to be around $1.0 \text{ mg L}^{-1}$ . This level provides significant dental benefits with minimal risk of adverse effects.

Harmful Effects of Excessive Fluoride:

When the concentration of fluoride in drinking water exceeds certain limits, it can become detrimental to health:

Dental Fluorosis: If the fluoride concentration is above approximately $1.5 \text{ mg L}^{-1}$ to $2.0 \text{ mg L}^{-1}$ , especially during tooth development in children, it can lead to dental fluorosis. This condition manifests as white spots, streaks, or even brown stains and pitting on the tooth enamel, making the teeth appear mottled.

Skeletal Fluorosis: At much higher concentrations, typically above $10 \text{ mg L}^{-1}$ , fluoride can accumulate in bones and joints over time, leading to skeletal fluorosis. This is a more severe condition characterized by pain, stiffness, and damage to bones and joints, potentially causing crippling deformities.

Determining the Maximum Safe Limit:

Considering both the beneficial effects and the potential for harm, regulatory bodies establish a maximum permissible limit for fluoride in drinking water. This limit aims to maximize the dental health benefits while preventing the onset of fluorosis. The generally accepted maximum limit for fluoride in drinking water to be considered safe is around $1.5 \text{ mg L}^{-1}$ to $2.0 \text{ mg L}^{-1}$ . Beyond this range, the risk of dental fluorosis increases significantly. …
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TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
$K_b = 0.52\, \text{K kg mol}^{-1}$
›Reveal solutionSolution
The boiling point elevation gives the molality of the solution; using that molality with the freezing point depression constant yields the freezing point depression, which is subtracted from the normal freezing point to get 272.964 K — option (C).

Concept & Intuition

Colligative properties depend only on the number of solute particles, not their identity. Here, boiling point elevation and freezing point depression are both proportional to the same molality of glucose. So we can find the molality from the given boiling point change, then use it to compute the freezing point change. The key is that the same solution has the same molality for both properties.

Step-by-step solution

Find the boiling point elevation Pure water boils at 373.15 K; the solution boils at 373.202 K.

$\Delta T_b = 373.202 - 373.15 = 0.052\ \text{K}$

Relate elevation to molality The formula is $\Delta T_b = K_b \cdot m$ , where $K_b = 0.52\ \text{K kg mol}^{-1}$ .

$m = \frac{\Delta T_b}{K_b} = \frac{0.052}{0.52} = 0.1\ \text{mol/kg}$

Use the same molality for freezing point depression For water, $K_f = 1.86\ \text{K kg mol}^{-1}$ .

$\Delta T_f = K_f \cdot m = 1.86 \times 0.1 = 0.186\ \text{K}$

Compute the freezing point …
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$^{-1}$
›Reveal solutionSolution

Relative lowering of vapour pressure gives the solute's mole fraction; solving for it yields molar mass $\approx 340\ \text{g mol}^{-1}$ (option C).

For a dilute solution, Raoult's law gives the relative lowering of vapour pressure equal to the mole fraction of the solute:
$\frac{p^{\circ}-p}{p^{\circ}} = \frac{n_{2}}{n_{1}}$
Given: $p^{\circ}=3.78\ \text{kPa}$ , $p=3.768\ \text{kPa}$ , so $p^{\circ}-p = 0.012\ \text{kPa}$ .

Moles of water: $n_{1} = \dfrac{1000}{18} = 55.56\ \text{mol}$ .

Let the molar mass of the solute be $M$ . Mass of solute $= 0.06\ \text{kg} = 60\ \text{g}$ , so $n_{2} = \dfrac{60}{M}$ . …
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$X_5Y_8$
›Reveal solutionSolution
In a bcc lattice, corners contribute 1/8 atom each and the center contributes 1 full atom. With 5 of 8 corners occupied by X and 1 center occupied by Y, the ratio X:Y = 5/8 : 1 = 5:8, so the formula is X₅Y₈.

Concept & Intuition

A body-centered cubic (bcc) unit cell has atoms at all eight corners and one atom at the body center. Each corner atom is shared by 8 adjacent cells, so only 1/8 of each corner atom belongs to one unit cell. The body-center atom belongs entirely to that cell. When some corner atoms are missing, we simply count how many remain and multiply by 1/8. The center atom is unaffected. The formula is the simplest whole-number ratio of X (corners) to Y (center).

Step-by-step reasoning

Full bcc cell counts

In a perfect bcc cell:

Corners: 8 atoms × 1/8 = 1 atom

Center: 1 atom × 1 = 1 atom So the ratio is 1:1, giving formula XY.

Effect of missing corner atoms

The problem says three atoms are missing from the corners. That means only $8 - 3 = 5$ corner atoms remain.

Each remaining corner contributes 1/8 of an atom to the cell.

So the total number of X atoms in the cell = $5 \times \frac{1}{8} = \frac{5}{8}$ .

Y atoms at the center

The center atom (Y) is still present and fully belongs to the cell: 1 Y atom.

Ratio of X to Y

X : Y = $\frac{5}{8} : 1$ . …
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TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$Q. Assertion (A): Graphite is used as a dry lubricant in machines which run at high temperatures Reason (R): The layers of graphite slip one over the other when pressure is applied The correct option among the following is (A) (A) is true but (R) is false (B) (A) is false but (R) is true (C) (A) and (R) are true. (R) is the correct explanation of (A) (D) (A) and (R) are true, but (R) is not correct explanation of (A)$
›Reveal solutionSolution
Graphite’s layered structure lets its sheets slide easily, making it a good dry lubricant at high temperatures. Both Assertion and Reason are true, and the Reason correctly explains the Assertion, so the answer is (C).

Concept & Intuition

Graphite is a form of carbon where atoms are arranged in flat, hexagonal sheets (like chicken wire). Within each sheet, bonds are very strong (covalent). But between sheets, only weak forces (van der Waals) hold them together. When you press or rub graphite, these sheets can slide over each other like a deck of cards. That’s why it feels slippery — it reduces friction. And because it’s a solid (not an oil), it works even at high temperatures where liquid lubricants would evaporate or burn.

Step-by-step reasoning

Check the Assertion (A): “Graphite is used as a dry lubricant in machines which run at high temperatures.”

Graphite is indeed a solid lubricant. Unlike oil or grease, it doesn’t decompose or vaporize easily at high temperatures (it sublimes only above 3600 °C). So it’s ideal for hot environments like kilns, engines, or industrial presses.

Result: Assertion is true.

Check the Reason (R): “The layers of graphite slip one over the other when pressure is applied.”

This is the key property: weak interlayer forces allow sliding. Even gentle pressure can make the layers shift, reducing friction.

Result: Reason is true.

Determine if (R) correctly explains (A): …
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TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$\left[ \mathrm{H_3O^+} \right] = 7.5 \times 10^{-8} \mathrm{M}$
›Reveal solutionSolution
The key idea is that in water at 25 °C, the ion‑product constant $K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$ . Given $[ \mathrm{H_3O^+} ] = 7.5 \times 10^{-8}\,\mathrm{M}$ , we solve for $[ \mathrm{OH^-} ]$ and get $1.3 \times 10^{-7}\,\mathrm{M}$ , which corresponds to option (B).

The relevant concept is the autoionization of water. Water molecules very slightly dissociate into hydronium and hydroxide ions, and at 25 °C the product of their concentrations is always $1.0 \times 10^{-14}$ . This means if you know one concentration, you can always find the other — no matter whether the solution is acidic, basic, or neutral.

A common pitfall is to assume that because the given $[ \mathrm{H_3O^+} ]$ is less than $1.0 \times 10^{-7}\,\mathrm{M}$ , the solution is basic, but then to forget that $[ \mathrm{OH^-} ]$ must be greater than $1.0 \times 10^{-7}\,\mathrm{M}$ — and to pick a tiny number like $2 \times 10^{-11}$ by mistake. Let’s avoid that.

Write the ion‑product expression. For water at 25 °C:

$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$

Substitute the known value. We are told $[ \mathrm{H_3O^+} ] = 7.5 \times 10^{-8}\,\mathrm{M}$ . So:

$(7.5 \times 10^{-8}) \times [\mathrm{OH^-}] = 1.0 \times 10^{-14}$

Solve for $[ \mathrm{OH^-} ]$ . Divide both sides by $7.5 \times 10^{-8}$ :

$[\mathrm{OH^-}] = \frac{1.0 \times 10^{-14}}{7.5 \times 10^{-8}}$

Do the arithmetic. First divide the coefficients: $1.0 / 7.5 \approx 0.1333$ . Then subtract exponents: $10^{-14} / 10^{-8} = 10^{-6}$ . …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$\mathrm{O_2}$
›Reveal solutionSolution
Henry's Law is valid for gases that dissolve physically without significant chemical reaction with the solvent. It applies to $\mathrm{O_2}$ dissolution in water and, for the dissolved molecular species, to $\mathrm{CO_2}$ dissolution in water, making option (C) the correct choice.

Concept and Intuition: The Essence of Henry's Law

Imagine a fizzy drink. When you open the bottle, bubbles escape. Why? Because the pressure of carbon dioxide above the liquid drops, and the gas rushes out of the solution. This everyday phenomenon illustrates Henry's Law: the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.

The Core Idea:

Henry's Law describes a dynamic equilibrium. Gas molecules are constantly moving between the gas phase above the liquid and the dissolved phase within the liquid. At equilibrium, the rate at which gas molecules enter the liquid equals the rate at which they leave. If you increase the partial pressure of the gas, you force more gas molecules into the liquid, increasing its concentration until a new equilibrium is established. This relationship is linear, meaning if you double the pressure, you double the dissolved concentration.

[!FORMULA]

Mathematically, Henry's Law is often expressed as:

$P = K_H c$

where:

$P$ is the partial pressure of the gas above the solution.

$c$ is the concentration (e.g., molarity or mole fraction) of the dissolved gas.

$K_H$ is Henry's Law constant, a unique value for a specific gas-solvent pair at a given temperature.

When Does It Work? (Conditions for Validity)

Henry's Law is an ideal gas-solution law, meaning it works best under specific conditions:

Low Partial Pressure: The gas pressure shouldn't be excessively high.

Low Concentration: The gas should be sparingly soluble in the liquid.

No Chemical Reaction: Crucially, the dissolved gas should not react chemically with the solvent or any other components in the solution. If it reacts, the dissolved species is no longer just the original gas molecule, and the simple proportionality breaks down.

No Dissociation/Association: The gas molecules should not break apart into ions (dissociate) or clump together (associate) in the solution.

Constant Temperature: The Henry's Law constant ( $K_H$ ) is highly dependent on temperature.

Step-by-Step Analysis of Each Option

Let's evaluate each scenario against these conditions, particularly focusing on the "no chemical reaction" rule.

Analyze Option A: Ammonia gas dissolution in water ( $\mathrm{NH_3}$ in $\mathrm{H_2O}$ )

Ammonia is a basic gas that reacts quite strongly with water to form ammonium hydroxide, which then partially dissociates: $\mathrm{NH_3(g) + H_2O(l) \rightleftharpoons NH_4OH(aq) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$

Because ammonia undergoes a significant chemical reaction and dissociation in water, its dissolution is not a simple physical process. The total concentration of dissolved ammonia species is not directly proportional to the partial pressure of $\mathrm{NH_3}$ in the straightforward manner described by Henry's Law.

Conclusion for A: Not valid.

Analyze Option B: $\mathrm{O_2}$ gas dissolution in unsaturated blood

Blood is a complex mixture, and its most significant component for oxygen transport is hemoglobin (Hb). Hemoglobin binds reversibly with oxygen: $\mathrm{Hb(aq) + O_2(g) \rightleftharpoons HbO_2(aq)}$

This is a vital chemical reaction (complex formation) that allows blood to carry far more oxygen than would be possible through simple physical dissolution in water alone. The binding of $\mathrm{O_2}$ to hemoglobin is a cooperative and complex process, and the total oxygen content in blood does not follow the simple linear relationship of Henry's Law.

Conclusion for B: Not valid.

Analyze Option C: $\mathrm{O_2}$ dissolution in water

Oxygen gas ( $\mathrm{O_2}$ ) is relatively unreactive and does not undergo significant chemical reactions with water. It dissolves physically in water.

At typical atmospheric pressures and temperatures, oxygen is sparingly soluble in water, meaning its concentration in solution is low.

These conditions align perfectly with the requirements for Henry's Law. Therefore, the dissolution of $\mathrm{O_2}$ in pure water is a classic and valid application of Henry's Law.

Conclusion for C: Valid.

Analyze Option D: $\mathrm{CO_2}$ dissolution in water

Carbon dioxide ( $\mathrm{CO_2}$ ) dissolves in water. The initial step is a physical dissolution: $\mathrm{CO_2(g) \rightleftharpoons CO_2(aq)}$ …
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TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
$Q. 1 mole of a real gas is kept at high pressure of 100 bar at 300 K. If van der Waals constant b is 0.005 L / mol, what are the values of compressibility factor Z of the gas and % deviation of volume from ideality? (A) Z % Deviation 1.10 10 (B) Z % Deviation 1.2 20 (C) Z % Deviation 1.02 2 (D) Z % Deviation 1.2 15$
›Reveal solutionSolution
At high pressure, the van der Waals equation simplifies to $P(V_m - b) = RT$ , so the compressibility factor $Z = 1 + \frac{Pb}{RT}$ . For the given data, $Z = 1.02$ and the % deviation of volume from ideality is $2\%$ , matching option (C).

The key to this problem is recognising what "high pressure" does to the van der Waals equation. At low pressures, the $a/V^2$ correction for intermolecular attraction matters a lot. But at high pressures, the molecules are already so close together that the attractive term becomes negligible compared to the repulsive volume correction $b$ . The gas behaves more like a hard-sphere gas where the only non-ideality comes from the finite size of molecules.

So at high pressure, the van der Waals equation $\left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT$ simplifies to $P(V_m - b) = RT$ . This is the starting point.

Find the molar volume of the real gas. From $P(V_m - b) = RT$ , we get $V_m = \frac{RT}{P} + b$ . Plug in the numbers: $R = 0.0821 \text{ L·bar·mol}^{-1}\text{K}^{-1}$ , $T = 300 \text{ K}$ , $P = 100 \text{ bar}$ , $b = 0.005 \text{ L/mol}$ .

$V_m = \frac{0.0821 \times 300}{100} + 0.005 = \frac{24.63}{100} + 0.005 = 0.2463 + 0.005 = 0.2513 \text{ L/mol}$

Find the ideal molar volume at the same $P$ and $T$ .

For an ideal gas, $V_m^{\text{ideal}} = \frac{RT}{P} = 0.2463 \text{ L/mol}$ .

Calculate the compressibility factor $Z$ .

$Z = \frac{PV_m}{RT} = \frac{V_m}{V_m^{\text{ideal}}}$ .

$Z = \frac{0.2513}{0.2463} \approx 1.0203 \approx 1.02$
Tip

You can also get $Z$ directly without finding $V_m$ : from $P(V_m - b) = RT$ , divide both sides by $RT$ to get $\frac{PV_m}{RT} - \frac{Pb}{RT} = 1$ , so $Z = 1 + \frac{Pb}{RT}$ . …
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TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
$Q. Assertion (A): Hydrogen fluoride has higher boiling point than other hydrogen halides. Reason (R): Hydrogen fluoride exhibits strong hydrogen bonding. The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true$
›Reveal solutionSolution
Hydrogen fluoride has an exceptionally high boiling point compared to other hydrogen halides because it forms strong intermolecular hydrogen bonds, which require more energy to overcome during boiling. Both the assertion and the reason are true, and the reason correctly explains the assertion.

The boiling point of a substance is determined by the strength of the intermolecular forces (IMFs) between its molecules. Stronger IMFs require more energy to overcome, leading to a higher boiling point. For simple molecular compounds, the primary IMFs are London dispersion forces, dipole-dipole interactions, and hydrogen bonding.

Evaluate Assertion (A): Hydrogen fluoride has higher boiling point than other hydrogen halides.

The hydrogen halides are $\text{HF}$ , $\text{HCl}$ , $\text{HBr}$ , and $\text{HI}$ . Generally, as we move down a group, the molecular size and number of electrons increase, leading to stronger London dispersion forces and thus higher boiling points.

Let's look at the approximate boiling points:

$\text{HF}$ : $19.5 \,^\circ\text{C}$

$\text{HCl}$ : $-85 \,^\circ\text{C}$

$\text{HBr}$ : $-66.8 \,^\circ\text{C}$

$\text{HI}$ : $-35.3 \,^\circ\text{C}$ If we follow the trend of increasing London dispersion forces, we would expect $\text{HF}$ to have the lowest boiling point among the hydrogen halides due to its smallest size. However, $\text{HF}$ has an anomalously high boiling point, significantly higher than $\text{HCl}$ , $\text{HBr}$ , and even $\text{HI}$ . This makes Assertion (A) true.

Evaluate Reason (R): Hydrogen fluoride exhibits strong hydrogen bonding.

Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (like fluorine, oxygen, or nitrogen) and is attracted to another highly electronegative atom in an adjacent molecule.

Fluorine is the most electronegative element in the periodic table. In the $\text{H-F}$ molecule, the fluorine atom pulls electron density strongly away from the hydrogen atom, creating a highly polar bond ( $\delta^+\text{H}-\text{F}\delta^-$ ). This leaves the hydrogen atom with a significant partial positive charge and a very small size, allowing it to form strong electrostatic attractions (hydrogen bonds) with the lone pairs of electrons on the fluorine atom of an adjacent $\text{HF}$ molecule. …
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TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$^\circ$
›Reveal solutionSolution
The pH of pure water decreases as temperature increases because the autoionization constant $K_w$ rises with temperature, making $[H^+]$ larger than $10^{-7}$ M; at 80 °C, pH < 7.0.

The key concept is that pH = –log[H⁺], and for pure water $[H^+] = \sqrt{K_w}$ . The value of $K_w$ is not constant — it increases with temperature. At 25 °C, $K_w = 1.0 \times 10^{-14}$ and pH = 7.0. But at higher temperatures, the equilibrium $2H_2O \rightleftharpoons H_3O^+ + OH^-$ shifts right (endothermic), so $[H^+]$ rises, and pH falls below 7.0 — even though the water remains neutral.

Recall the definition of neutrality

Pure water is neutral when $[H^+] = [OH^-]$ . The ion product is $K_w = [H^+][OH^-]$ . For neutral water, $[H^+] = [OH^-] = \sqrt{K_w}$ .

Know how $K_w$ changes with temperature

The autoionization of water is endothermic ( $\Delta H > 0$ ). By Le Chatelier’s principle, raising temperature shifts the equilibrium to produce more ions. At 25 °C, $K_w = 1.0 \times 10^{-14}$ . At 80 °C, $K_w$ is roughly $2.5 \times 10^{-13}$ (a known experimental value).

Calculate $[H^+]$ at 80 °C

$[H^+] = \sqrt{K_w} = \sqrt{2.5 \times 10^{-13}} \approx 5.0 \times 10^{-7} \ \text{M}$

Find the pH
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Q.We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van't Hoff factor for these solutions will be in the order______.

Concept understanding — VanT Hoff Factor Association

The Intuition: What Happens When Particles Stick Together?

The Van't Hoff Factor: The Correction Number

A Concrete Example: Acetic Acid in Benzene

The General Formula for Association

How Association Affects Colligative Properties

Why this formula?

Van't Hoff Factor for Association: Why the Formula Holds

1. The Core Idea: What Changes?

2. Setting Up the Association Process

What happens to the particles?

3. Total Particles After Association

4. The Van't Hoff Factor Formula

5. Why This Makes Physical Sense

Why the van’t Hoff factor depends on concentration

Step-by-step reasoning

Method: Effect of Dilution on van’t Hoff Factor for Strong Electrolytes

Steps to determine the order:

Common Mistakes Students Make on van't Hoff Factor & Concentration

Mistake 1: Assuming $i$ is constant for all concentrations of the same solute

Mistake 2: Thinking higher concentration means higher $i$

Mistake 3: Picking option (iii) — $i_A = i_B = i_C$

Mistake 4: Picking option (ii) — $i_A > i_B > i_C$

Concept and Intuition: The Essence of Henry's Law

Step-by-Step Analysis of Each Option

Q.We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentrations 0.1M, 0.01M and 0.001M, respectively. The value of van't Hoff factor for these solutions will be in the order______.

Why the van’t Hoff factor depends on concentration

Step-by-step reasoning

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