Chemistry
PYQ Solutions

Osmosis: Osmosis is the spontaneous flow of solvent molecules through a semipermeable membrane from a region of lower solute concentration (higher solvent concentration) to a region of higher solute concentration (lower solvent concentration).

Osmotic Pressure (π): The excess pressure that must be applied on the solution side to just stop the flow of solvent through the semipermeable membrane is called osmotic pressure.

where: C = molar concentration of solute, R = gas constant = 0.0821 L atm K⁻¹ mol⁻¹, T = temperature in Kelvin, n_B = moles of solute, V = volume of solution.

Is osmotic pressure a colligative property? → YES

Justification: Osmotic pressure depends only on the number of solute particles (moles of solute per unit volume) and NOT on the nature or identity of the solute. This is the defining characteristic of a colligative property — one that depends on the number of dissolved particles, not their type.

For example, 0.1 M glucose and 0.1 M sucrose exert the same osmotic pressure because they have the same molar concentration (same number of particles per litre).

Colligative Properties: Properties of solutions that depend only on the number of solute particles dissolved in a definite amount of solvent, and NOT on the nature, size, or chemical identity of the solute particles.

The word "colligative" comes from the Latin colligatus meaning "bound together", indicating that these properties are collectively determined by the number of particles.

Four Colligative Properties:

Key Point: Colligative properties are observed only for non-volatile solutes dissolved in volatile solvents (for vapour pressure related properties).

Reverse Osmosis: When a pressure greater than the osmotic pressure is applied on the solution side, the direction of solvent flow is reversed — solvent flows from the solution to the pure solvent side. This is called Reverse Osmosis (RO).

Normal osmosis: Pure solvent → Solution
Reverse osmosis: Solution → Pure solvent (when P > π)

Condition: The external pressure applied must be greater than the osmotic pressure of the solution: \(P_{applied} > \pi_{solution}\)

One Important Use: Reverse osmosis is used for the desalination of sea water to obtain pure, drinkable water. Sea water is passed through a RO membrane (made of polyamide) under very high pressure (~30 atm), and pure water filters through while dissolved salts are rejected.

Henry's Law: At constant temperature, the partial pressure of a gas in the vapour phase is directly proportional to the mole fraction of the gas dissolved in the liquid.

where p = partial pressure, x = mole fraction of gas in solution, K_H = Henry's law constant. Higher K_H → lower solubility at a given pressure.

Effect of Temperature: The solubility of a gas in a liquid decreases with increase in temperature. Dissolution of gas is exothermic: $$\text{Gas} + \text{Solvent} \rightleftharpoons \text{Solution} + \text{Heat}$$ By Le Chatelier's principle, increasing temperature shifts equilibrium backward → solubility decreases.

Why do fish prefer cold water?

• Oxygen (O₂) is essential for survival of fish.
• Solubility of O₂ in water is higher at lower temperatures.
• Cold water contains more dissolved oxygen, making it easier for fish to breathe through gills.
• In warm water, O₂ content is lower, making survival difficult.

Molality (m): Number of moles of solute dissolved per kilogram (1000 g) of solvent.

Units: mol kg⁻¹ (or molal)

Conclusion: Molality is preferred over molarity for colligative property calculations and thermodynamic studies because it is temperature-independent.

In 100 g of solution: Mass of benzene (C₆H₆) = 30 g, Mass of CCl₄ = 70 g. M(benzene) = 78 g/mol, M(CCl₄) = 154 g/mol

Step 1: \(n_{\text{benzene}} = \frac{30}{78} = 0.3846 \text{ mol}\)

Step 2: \(n_{\text{CCl}_4} = \frac{70}{154} = 0.4545 \text{ mol}\)

Step 3: \(n_{\text{total}} = 0.3846 + 0.4545 = 0.8391 \text{ mol}\)

Van't Hoff Factor (i):

Relation with Degree of Dissociation (α): For electrolyte dissociating into n ions: $$i = 1 + (n - 1)\alpha \quad\therefore \alpha = \frac{i - 1}{n - 1}$$ Example: NaCl → Na⁺ + Cl⁻ → n = 2 → i = 1 + α

Relation with Degree of Association (α): For n molecules combining into 1: $$i = 1 - \alpha\left(1 - \frac{1}{n}\right) \quad\therefore \alpha = \frac{(1 - i)}{\left(1 - \frac{1}{n}\right)}$$ Example: Acetic acid dimerises in benzene (n = 2) → i = 1 − α/2

Summary: i > 1 for dissociation; i < 1 for association; i = 1 for non-electrolytes.

Answer: Negative Deviation from Raoult's Law

Reasoning: When phenol and aniline are mixed, a new and stronger intermolecular hydrogen bond is formed between the —OH group of phenol and the —NH₂ group of aniline: \(\text{C}_6\text{H}_5\text{—OH} \cdots \text{NH}_2\text{—C}_6\text{H}_5\). This interaction is stronger than interactions in either pure liquid.

• Tendency of molecules to escape from solution (vapour pressure) is reduced
• \(P_{\text{total}} < \chi_A P_A^\circ + \chi_B P_B^\circ\)
• ΔH_mix < 0 (exothermic mixing)
• ΔV_mix < 0 (volume decreases on mixing)

(i) Mole Fraction: The ratio of the number of moles of that component to the total number of moles of all components. $$\chi_B = \frac{n_B}{n_A + n_B}$$

Example: 1 mol NaCl in 9 mol water: \(\chi_{\text{NaCl}} = \frac{1}{10} = 0.1\). Note: Sum of all mole fractions = 1.

(ii) Isotonic Solutions: Two solutions are isotonic if they have the same osmotic pressure at the same temperature — equal molar concentrations. No net flow of solvent between them when separated by a semipermeable membrane. $$\pi_1 = \pi_2 \Rightarrow C_1 = C_2$$

Example: 0.9% (w/v) NaCl (normal saline) is isotonic with blood plasma — used in intravenous drips to avoid osmotic shock.

Raoult's Law: The partial vapour pressure of each component is directly proportional to its mole fraction in the solution.

Two Conditions for Ideal Behaviour:

1. Solute–Solvent interactions (A−B) must equal Solute–Solute (B−B) and Solvent–Solvent (A−A) interactions → ensures ΔH_mix = 0
2. Volume of solution = sum of volumes of pure components → ΔV_mix = 0 (no expansion or contraction)

Examples of ideal solutions: Benzene + toluene; n-hexane + n-heptane; ethyl bromide + ethyl iodide.

Positive Deviation: \(P_{\text{observed}} > \chi_A P_A^\circ + \chi_B P_B^\circ\). A−B interactions are WEAKER than A−A and B−B. Molecules escape more easily → higher VP. ΔH_mix > 0 (endothermic), ΔV_mix > 0. Example: Ethanol + Water, CCl₄ + Toluene.

Negative Deviation: \(P_{\text{observed}} < \chi_A P_A^\circ + \chi_B P_B^\circ\). A−B interactions are STRONGER. Molecules escape less easily → lower VP. ΔH_mix < 0 (exothermic), ΔV_mix < 0. Example: Chloroform + Acetone, Phenol + Aniline.

M(glucose, C₆H₁₂O₆) = 180 g/mol, Mass of water = 100 g

\(n_{\text{glucose}} = \frac{1.8}{180} = 0.01 \text{ mol}\), \(n_{\text{water}} = \frac{100}{18} = 5.556 \text{ mol}\)

Answer: Positive Deviation from Raoult's Law

Reason: In pure ethanol: strong hydrogen bonds (−OH group). In pure acetone: dipole–dipole interactions (C=O group). When mixed: the −OH of ethanol interacts with C=O of acetone, but this new interaction is weaker than original H-bonds in pure ethanol → higher vapour pressure → positive deviation.

Intermolecular Forces:

• Broken: H-bonds between ethanol molecules; dipole–dipole between acetone molecules
• Formed: Weaker dipole–dipole/H-bond interactions between ethanol and acetone (O−H···O=C)

Forces broken > forces formed (in strength): ΔH_mix > 0 (endothermic)

Ideal Solution: A solution that obeys Raoult's law exactly over the entire range of concentrations and at all temperatures. ΔH_mix = 0, ΔV_mix = 0.

Examples: Benzene and toluene; n-hexane and n-heptane; Ethyl bromide and ethyl iodide.

Two Conditions:

1. The intermolecular interactions between A−B molecules must be similar in magnitude to A−A and B−B interactions (same type of forces) → ΔH_mix = 0.
2. The components must be chemically similar (similar structure, polarity, size) → ΔV_mix = 0.

HCl in Benzene → Positive Deviation: Benzene has London dispersion forces (weak); HCl has dipole–dipole (stronger). When mixed, HCl + benzene interaction (weak induced dipole) is weaker than both originals → higher VP → positive deviation.

Chloroform + Acetone → Negative Deviation: H atom of CHCl₃ (δ+) forms a hydrogen bond with O of acetone (δ−): CHCl₃···O=C(CH₃)₂. This new interaction is stronger than original interactions → reduced VP → negative deviation.

(a) Aquatic life in cold water: Solubility of O₂ is governed by Henry's law. At lower temperatures, K_H is smaller → higher solubility of O₂. Cold water contains more dissolved O₂, essential for respiration of aquatic life. As water warms, O₂ content drops → aquatic life is stressed.

(b) Osmotic pressure preferred for high molecular mass polymers:

• Sensitivity: π = CRT → even a very small concentration gives a measurable pressure.
• Room temperature measurements are possible (no need to heat/cool) → no decomposition of thermally sensitive polymers.
• ΔT_b and ΔT_f are extremely small for high MW solutes (too small to measure accurately), but π remains practically measurable.

\(n_{\text{urea}} = \frac{25}{60} = 0.4167 \text{ mol}\), mass of water = 0.5 kg

Molality: \(m = \frac{0.4167}{0.5} = 0.8333 \text{ mol kg}^{-1}\)

Elevation: \(\Delta T_b = K_b \times m = 0.52 \times 0.8333 = 0.4333 \text{ K}\)

Azeotrope: A mixture of two liquids with a constant boiling point and constant composition throughout distillation — both liquid and vapour phases have identical compositions. Cannot be separated by simple fractional distillation.

Minimum Boiling Azeotrope: Formed by solutions showing positive deviation. B.P. is lower than both pure components. Example: Ethanol (b.p. 78.5°C) + Water (b.p. 100°C) → Azeotrope at 78.1°C (95.5% ethanol).

Maximum Boiling Azeotrope: Formed by solutions showing negative deviation. B.P. is higher than both pure components. Example: HNO₃ (b.p. 86°C) + Water (b.p. 100°C) → Azeotrope at 120.5°C (68% HNO₃).

χ_ethanol = 0.040, χ_water = 0.960. Consider 1 mol of solution: n_ethanol = 0.040 mol, n_water = 0.960 mol.

Mass of water = 0.960 × 18 = 17.28 g = 0.01728 kg

Mass of solution = (0.040 × 46) + (0.960 × 18) = 19.12 g; V = 19.12/0.997 = 19.18 mL = 0.01918 L

Abnormal Molar Mass: When experimentally determined molar mass using colligative properties differs from the expected (theoretical) molar mass. Occurs when solute undergoes dissociation or association. $$M_{\text{observed}} = \frac{M_{\text{theoretical}}}{i}$$

Van't Hoff Factor (i): $$i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}} = \frac{M_{\text{theoretical}}}{M_{\text{observed}}}$$

When i > 1: Solute dissociates → more particles → higher colligative property. Example: NaCl → Na⁺ + Cl⁻ → i ≈ 2.

When i < 1: Solute undergoes association (molecules cluster). Example: CH₃COOH dimerises in benzene: 2CH₃COOH ⇌ (CH₃COOH)₂ → fewer particles → i < 1.

Given: 5% sucrose (M = 342) freezes at 271 K → ΔT_f(sucrose) = 273.15 − 271 = 2.15 K. Kf = 1.86 K kg mol⁻¹.

Molality of 5% glucose: In 100 g solution: 5 g glucose, 95 g water = 0.095 kg.

\(m_{\text{glucose}} = \frac{5/180}{0.095} = \frac{0.02778}{0.095} = 0.2924 \text{ mol/kg}\)

ΔT_f for glucose: \(\Delta T_f = K_f \times m = 1.86 \times 0.2924 = 0.5439 \text{ K}\)

\(n = \frac{18}{180} = 0.1 \text{ mol}\), m = 0.1/1 = 0.1 mol/kg

\(\Delta T_b = 0.52 \times 0.1 = 0.052 \text{ K}\)

KNO₃ → K⁺ + NO₃⁻ → i = 2, M(KNO₃) = 101 g/mol.

Using \(\Delta T_f = i \times K_f \times m\): \(2 = 2 \times 1.86 \times m \Rightarrow m = \frac{2}{3.72} = 0.5376 \text{ mol/kg}\)

Using Raoult's Law: \(600 = \chi_A \times 450 + (1 - \chi_A) \times 700\)

\(600 = 450\chi_A + 700 - 700\chi_A \Rightarrow -100 = -250\chi_A \Rightarrow \chi_A = 0.4, \chi_B = 0.6\)

Liquid phase: χ_A = 0.4, χ_B = 0.6

Partial pressures: p_A = 0.4 × 450 = 180 mmHg; p_B = 0.6 × 700 = 420 mmHg

Vapour phase: y_A = 0.3, y_B = 0.7

P = 0.80 P° → \(\frac{P^\circ - P}{P^\circ} = 0.20 = \chi_{\text{solute}}\). n_octane = 114/114 = 1 mol.

\(\frac{n_s}{n_s + 1} = 0.20 \Rightarrow n_s = 0.25 \text{ mol}\)

ΔT_b = 0.42 K, W_water = 0.5 kg.

m = 0.42/0.512 = 0.8203 mol/kg; n = 0.8203 × 0.5 = 0.4102 mol

ΔT_b (observed) = 0.18°C. \(m_{\text{obs}} = \frac{0.18}{0.512} = 0.3516 \text{ mol/kg}\)

Van't Hoff factor: \(i = \frac{0.3516}{1.00} = 0.3516\)

For dimerisation (n=2): \(i = 1 - \frac{\alpha}{2} \Rightarrow \alpha = 2(1-i) = 2(1-0.3516) = 2 \times 0.6484\)

Note: The standard CBSE answer gives degree of association ≈ 64.8% ≈ 65%.

n = 1.0/185000 = 5.405 × 10⁻⁶ mol; V = 4.5 × 10⁻⁴ m³; T = 310 K.

C = 5.405×10⁻⁶ / 4.5×10⁻⁴ = 1.201×10⁻² mol m⁻³

n_glucose = 10/180 = 0.0556 mol; n_water = 90/18 = 5 mol.

\(\chi_{\text{glucose}} = \frac{0.0556}{5.0556} = 0.011\)

W = 5 g; V = 0.5 L; π = 4.26 bar; T = 300 K. Using π = (W/MV)RT:

P(CO₂) = 2.5 × 101325 = 2.533 × 10⁵ Pa. n_water = 500/18 = 27.78 mol.

\(\chi_{\text{CO}_2} = \frac{p}{K_H} = \frac{2.533 \times 10^5}{1.67 \times 10^8} = 1.517 \times 10^{-3}\)

\(n_{\text{CO}_2} \approx \chi_{\text{CO}_2} \times n_{\text{water}} = 1.517 \times 10^{-3} \times 27.78 = 0.04214 \text{ mol}\)

m = (68.5/342)/1 = 0.2002 mol/kg.

C = 0.2002/1.0685 = 0.1874 mol/L; π = 0.1874 × 0.0821 × 300 = 4.615 atm.

TYPE 1: Positive Deviation — P_observed > P_Raoult. A−B interactions are weaker than A−A and B−B. Molecules are held less tightly → escape more easily → higher VP. ΔH_mix > 0 (endothermic), ΔV_mix > 0.

Examples: (1) Ethanol + Water: When mixed, H-bonds between unlike molecules are weaker → positive deviation. (2) CCl₄ + Toluene: Unlike molecules have weaker interactions. (3) Acetone + CS₂.

TYPE 2: Negative Deviation — P_observed < P_Raoult. A−B interactions are stronger than A−A and B−B. Molecules held more tightly → lower VP. ΔH_mix < 0 (exothermic), ΔV_mix < 0.

Examples: (1) Chloroform + Acetone: CHCl₃ (δ+ H) H-bonds with C=O of acetone. (2) Phenol + Aniline: O−H of phenol H-bonds with N−H of aniline. (3) HNO₃ + Water: Strong acid–base interaction → exothermic.

K₂SO₄ → 2K⁺ + SO₄²⁻ → i = 3. n = 0.025/174 = 1.437×10⁻⁴ mol; C = 7.184×10⁻⁵ mol/L.