NEET Solutions Previous Year Questions with Solutions

An ideal solution is defined as one that obeys Raoult’s law over the entire range of concentration. For such solutions: 1. Enthalpy of mixing is zero (Δ_mix H = 0), meaning no heat is absorbed or evolved. 2. Volume of mixing is zero (Δ_mix V = 0), meaning the total volume equals the sum of the individual volumes.

• • Assertion A is scientifically correct.

In liquid solutions, intermolecular cohesive forces (such as dipole-dipole, London dispersion, or van der Waals forces) are always present. An ideal solution does not mean "zero interaction". Rather, it means that the magnitude of solute-solvent (P-Q) intermolecular interactions is EQUAL to solute-solute (P-P) and solvent-solvent (Q-Q) interactions.

• • Because the bond-breaking energy of P-P and Q-Q equals the bond-making energy of P-Q, Δ_mix H = 0.
• • Reason R claims "No interactions occur", which is scientifically false.

Since Assertion A is a correct factual statement and Reason R is scientifically false, the correct choice is: A is correct but R is not correct.

• • Matches Option (3).

An ideal solution is defined as a binary solution in which each component obeys Raoult's law over the entire range of concentration and temperature: P_A = P_A° x_A and P_B = P_B° x_B.

• • Statement A is correct.

Chloroform (CHCl₃) and acetone (CH₃COCH₃) molecules form intermolecular hydrogen bonds between the acidic hydrogen of chloroform and the carbonyl oxygen of acetone. Because A-B interactions are STRONGER than A-A and B-B interactions, vapor escaping tendency decreases, leading to lower vapor pressure than predicted by Raoult's law (Negative deviation).

• • Δ_mix H < 0 and Δ_mix V < 0.
• • Statement B is correct.

Phenol and aniline form strong intermolecular hydrogen bonds between the phenolic proton (-OH) and the lone pair on the nitrogen atom of aniline (-NH₂). This A-B interaction is significantly stronger than the individual intermolecular interactions. Consequently, the mixture shows NEGATIVE deviation from Raoult's law, NOT positive deviation.

• • Statement C is incorrect.

Statements A and B only are correct.

• • Matches Option (1).

Examine solutions where the dispersion medium (solvent) is a gas:

• • A matches II (A-II).
• • Smoke: Carbon soot particles (solid) dispersed in air (gas) represents Solid in gas (III).
• • D matches III (D-III).

Examine solutions where the solvent is a solid:

• • Alloys (e.g., brass, bronze, gold-copper) consist of solid solute dissolved in solid solvent (B-I).
• • Amalgams consist of liquid mercury (solute) dissolved in solid metal like sodium or zinc (Liquid in solid, C-IV).

Assemble the pairs into the final selection:

• • Matches Option (2) exactly as specified in NCERT Table 2.1.

The elevation in boiling point ΔT_b of an electrolyte solution depends on the total concentration of dissolved solute particles:

• • For dilute aqueous solutions, molality m is approximately equal to molarity M.
• • The solution exhibiting the greatest elevation ΔT_b will have the highest boiling point T_b.

Assuming complete dissociation for strong electrolytes:

• • (1) Urea: Non-electrolyte ⟹ i = 1 ⟹ i × M = 0.01 M.
• • (2) KNO₃: Dissociates into K⁺ + NO₃⁻ ⟹ i = 2 ⟹ i × M = 2 × 0.01 = 0.02 M.
• • (3) Na₂SO₄: Dissociates into 2 Na⁺ + SO₄²⁻ ⟹ i = 3 ⟹ i × M = 3 × 0.01 = 0.03 M.
• • (4) Glucose: Non-electrolyte ⟹ i = 1 ⟹ i × M = 1 × 0.015 = 0.015 M.

Rank solutions by decreasing effective particle concentration:

• • 0.01 M Na₂SO₄ produces the greatest particle concentration and thus the highest boiling point.
• • Matches Option (3) with complete precision.

Total moles in the binary liquid solution:

• • x_X = 5 / 15 = 1/3.
• • x_Y = 10 / 15 = 2/3.

If the solution were ideal, the total vapor pressure would follow Raoult’s law:

• • The ideal expected total pressure is 73 torr.

Evaluate the direction of deviation:

• • Because the observed vapor pressure is lower than predicted by Raoult’s law, escaping tendency is suppressed due to stronger intermolecular attractions (X-Y > X-X, Y-Y).
• • This defines NEGATIVE DEVIATION from Raoult’s law, matching Option (2).

Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution:

From the rearranged equation, at a given constant pressure p, the mole fraction x (which measures gas solubility) is inversely proportional to the Henry’s law constant KH: Higher KH -> Lower solubility Lower KH -> Higher solubility

Listing the given KH values in increasing numerical order: - Gas B: KH = 2 × 10⁻⁵ kbar (exceptionally small KH -> maximum solubility) - Gas C: KH = 35 kbar (intermediate KH -> moderate solubility) - Gas A: KH = 145 kbar (largest KH -> minimum solubility)

For a dilute solution of a non-electrolyte, the osmotic pressure Π is directly proportional to the molar concentration C and absolute temperature T:

Plotting Π on the y-axis against concentration C on the x-axis gives a straight line passing through the origin: y = mx ⟹ Π = (RT) · C. Therefore, the slope of the line equals the product of the gas constant R and absolute temperature T:

Given Slope = 25.73 L bar mol⁻¹ and R = 0.083 L bar mol⁻¹ K⁻¹:

Converting the temperature from Kelvin to degrees Celsius:

Under high hydrostatic pressure deep underwater, atmospheric gases (especially nitrogen) dissolve in the diver’s blood in large amounts in accordance with Henry’s Law (p = K_H · x). When the diver ascends towards the surface, the ambient pressure decreases, and the dissolved gases are released from solution. Nitrogen forms gas bubbles in the bloodstream, blocking capillaries and causing a severe, painful medical condition known as the bends.

To prevent the formation of nitrogen bubbles and avoid nitrogen narcosis, commercial deep-sea diving breathing mixtures are diluted with helium (typically 11.7% He, 56.2% N₂, and 32.1% O₂). Therefore, Assertion A is completely true.

Helium is utilized specifically because it has an exceptionally high Henry’s law constant (K_H) in water/blood, which corresponds to an extremely LOW SOLUBILITY IN BLOOD. Gases mix in all proportions in the gas phase; claiming that helium is used because it has "high solubility in O₂" is scientifically baseless and factually false.

Assertion A is true, but Reason R is false. This corresponds to Option (4).

For dilute non-electrolyte solutions that neither associate nor dissociate (van 't Hoff factor $i = 1$), osmotic pressure $\Pi$ is given by:

Each solution contains $w = 10\text{ g}$ of solute dissolved in $V = 250\text{ mL} = 0.25\text{ L}$ of water: - Solution 1 (Glucose, $C_6H_{12}O_6$): $M_1 = 180\text{ g/mol}$ $$n_1 = \frac{10}{180} = 0.0556\text{ mol} \implies C_1 = \frac{0.0556}{0.25} = 0.222\text{ M}$$ - Solution 2 (Urea, $NH_2CONH_2$): $M_2 = 60\text{ g/mol}$ $$n_2 = \frac{10}{60} = 0.1667\text{ mol} \implies C_2 = \frac{0.1667}{0.25} = 0.667\text{ M}$$ - Solution 3 (Sucrose, $C_{12}H_{22}O_{11}$): $M_3 = 342\text{ g/mol}$ $$n_3 = \frac{10}{342} = 0.0292\text{ mol} \implies C_3 = \frac{0.0292}{0.25} = 0.117\text{ M}$$

Because osmotic pressure is directly proportional to molar concentration $C$:

Let the number of moles of benzene ($B$) be $n_B = 3x$ and octane ($O$) be $n_O = 2x$. The total moles in the solution are $n_{\text{total}} = 3x + 2x = 5x$. - Mole fraction of benzene: $$\chi_B = \frac{n_B}{n_{\text{total}}} = \frac{3x}{5x} = \frac{3}{5} = 0.6$$ - Mole fraction of octane: $$\chi_O = \frac{n_O}{n_{\text{total}}} = \frac{2x}{5x} = \frac{2}{5} = 0.4$$

For an ideal solution, the partial vapour pressure of each volatile component is equal to the vapour pressure of the pure component multiplied by its mole fraction in the liquid phase: - Partial pressure of benzene: $$P_B = P_B^\circ \cdot \chi_B = 280\text{ mm Hg} \times 0.6 = 168\text{ mm Hg}$$ - Partial pressure of octane: $$P_O = P_O^\circ \cdot \chi_O = 420\text{ mm Hg} \times 0.4 = 168\text{ mm Hg}$$

Applying Dalton's Law of Partial Pressures:

Depression in freezing point ($$\Delta T_f$$) is a colligative property directly proportional to the molal concentration of the solute in the dilute solution: $$\Delta T_f = i \cdot K_f \cdot m$$

Because the solute is explicitly specified as a non-electrolyte, it undergoes neither association nor dissociation in benzene, so $$i = 1$$. - Cryoscopic constant ($$K_f$$) = $$5.12\text{ K kg mol}^{-1}$$ - Molality ($$m$$) = $$0.078\text{ m} = 0.078\text{ mol kg}^{-1}$$.

$$\Delta T_f = 5.12 \times 0.078$$ $$\Delta T_f = 0.39936\text{ K}$$

Rounding $$0.39936\text{ K}$$ to two decimal places gives: $$\Delta T_f \approx 0.40\text{ K}$$ This precisely matches Option (3).

In solutions exhibiting positive deviation from Raoult's law, the intermolecular attractive forces between solute and solvent molecules ($$A-B$$) are weaker than those between like molecules ($$A-A$$ and $$B-B$$). As a result, molecules escape more easily into the vapor phase, leading to higher vapor pressures ($$P_A > P_A^\circ x_A$$, $$P_B > P_B^\circ x_B$$), endothermic mixing ($$\Delta_{\text{mix}} H > 0$$), and volume expansion ($$\Delta_{\text{mix}} V > 0$$).

In pure ethanol, molecules are strongly associated via intermolecular hydrogen bonding. When acetone is mixed with ethanol, its molecules slip in between host ethanol molecules and disrupt their hydrogen-bonded network. Because the new ethanol-acetone dipole-dipole attractions are weaker than the original ethanol-ethanol hydrogen bonds, the escaping tendency of both components increases, displaying positive deviation.

- Benzene + Toluene: Both are non-polar hydrocarbons of similar size and structure, forming an nearly IDEAL solution ($$\Delta H = 0, \Delta V = 0$$). - Chloroethane + Bromoethane: Ideal solution of homologous haloalkanes. - Acetone + Chloroform: Chloroform forms a new, strong intermolecular hydrogen bond with the carbonyl oxygen of acetone, displaying NEGATIVE deviation.

Ethanol + Acetone uniquely exhibits positive deviation from Raoult's law.