CBSE Previous Year Questions Solutions - Chemistry Class 12 - NEET
CBSE Previous Year Questions 2025
Multiple Choice Type Questions
Q1: In case of association, abnormal molar mass of solute will [1 Mark]
(A) increase
(B) decrease
(C) remain same
(D) first increase and then decrease
Ans: (A) increase
Sol: Colligative properties, such as the elevation of boiling point or depression of freezing point, are inversely proportional to the molar mass of the solute. In the case of association (e.g., dimerization of acetic acid in benzene), the number of particles in the solution decreases.
Since the number of particles decreases, the observed colligative property decreases, which leads to an increase in the calculated (abnormal) molar mass of the solute.
Q2: An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because __________. [1 Mark]
(A) it gains water due to osmosis
(B) it loses water due to reverse osmosis
(C) it gains water due to reverse osmosis
(D) it loses water due to osmosis
Ans: (D) it loses water due to osmosis
Sol: When an unripe mango is placed in a concentrated salt solution (a hypertonic solution), the water concentration inside the mango is higher than that in the salt solution. Through the process of osmosis, water molecules move from the region of higher concentration (inside the mango) to the region of lower concentration (the salt solution) through the semi-permeable membrane of the mango. This loss of water causes the mango to shrivel.
Q3: The value of Henry’s constant \(K_H\) is: [1 Mark]
(A) greater for gases with higher solubility
(B) greater for gases with lower solubility
(C) constant for all gases
(D) not related to the solubility of gases
Ans: (B) greater for gases with lower solubility
Sol: According to Henry’s Law, the partial pressure of a gas in the vapour phase (\(p\)) is proportional to the mole fraction of the gas (\(x\)) in the solution: \(p = K_H \cdot x\). At a constant pressure, \(K_H\) is inversely proportional to the mole fraction (\(x\)). Therefore, higher the value of \(K_H\) at a given pressure, the lower is the solubility of the gas in the liquid.
Q4: Which of the following aqueous solutions will have the highest freezing point? [1 Mark]
(A) 1.0 M KCl
(B) 1.0 M \(Na_{2}SO_{4}\)
(C) 1.0 M Glucose
(D) 1.0 M \(AlCl_{3}\)
Ans: (C) 1.0 M Glucose
Sol: Freezing point is a colligative property. The depression in freezing point (\(\Delta T_f\)) is directly proportional to the number of particles (van't Hoff factor, \(i\)). The solution with the lowest number of particles will have the minimum depression, and therefore the highest freezing point.
- \(KCl \rightarrow K^+ + Cl^-\) (\(i = 2\))
- \(Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}\) (\(i = 3\))
- Glucose \(\rightarrow\) does not dissociate (\(i = 1\))
- \(AlCl_3 \rightarrow Al^{3+} + 3Cl^-\) (\(i = 4\))
Since glucose has the smallest \(i\) value, its freezing point is the highest.
Q5: Assertion (A): Henry's law constant \((K_{H})\) decreases with increase in temperature.
Reason (R): As the temperature increases, solubility of gases in liquids decreases. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Ans: (D) Assertion (A) is false, but Reason (R) is true.
Sol:
Step-wise Justification:
- Reason Check: It is a fact that the solubility of gases in liquids decreases as temperature increases because the dissolution of a gas is typically an exothermic process.
- Assertion Check: According to Henry's Law, \(p = K_H \cdot \chi\) (where \(\chi\) is solubility). Since solubility (\(\chi\)) decreases with temperature, the value of \(K_H\) must increase to maintain the same partial pressure. Thus, the assertion is false.
Q6: The freezing point of one molal KCl solution, assuming KCl to be completely dissociated in water, is:
\((K_f \text{ for water} = 1.86 \text{ K kg mol}^{-1})\) [1 Mark]
(A) \(-3.72°C\)
(B) \(+3.72°C\)
(C) \(-1.86°C\)
(D) \(+2.72°C\)
Ans: (A) \(-3.72°C\)
KCl dissociates completely: \(\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-\), so van't Hoff factor \(i = 2\).
\(\Delta T_f = i \times K_f \times m = 2 \times 1.86 \times 1 = 3.72 \text{ K}\)
Freezing point \(= 0 - 3.72 = -3.72°C\)
Q7: A solution of acetone in ethanol: [1 Mark]
(A) Obeys Raoult's law.
(B) Forms an ideal solution.
(C) Shows a positive deviation from Raoult's law.
(D) Shows a negative deviation from Raoult's law.
Ans: (C) Shows a positive deviation from Raoult's law.
Sol: Acetone–ethanol interactions are weaker than acetone–acetone and ethanol–ethanol interactions (ethanol has H-bonding which is disrupted), so the solution shows positive deviation from Raoult's law (vapour pressure is higher than ideal).
Q8: Which of the following solutions will have the highest boiling point in water? [1 Mark]
(A) 1% KCl
(B) 1% glucose
(C) 1% urea
(D) 1% CaCl2
Ans: (D) 1% CaCl2
Sol: Boiling point elevation depends on number of particles (ΔTᵦ = iKᵦm). CaCl₂ dissociates into 3 ions (i = 3), which is highest among the options.
Therefore, 1% CaCl₂ has the highest boiling point. (Option D)
Q9: Assertion (A): Cooking time is reduced in pressure cooker.
Reason (R): Boiling point of water inside the pressure cooker is elevated. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Ans: (A)
Sol: In a pressure cooker, pressure increases, which raises the boiling point of water. Food cooks at a higher temperature, so cooking time is reduced.
Hence, both A and R are true and R correctly explains A. (Option A)
Q10: Assertion (A): A mixture of o-nitrophenol and p-nitrophenol can be separated by fractional distillation.
Reason (R): o-nitrophenol is steam volatile due to intramolecular hydrogen bonding. [1 Mark]
(A) Both A and R are true; R is the correct explanation of A.
(B) Both A and R are true; R is NOT the correct explanation of A.
(C) A is true, R is false.
(D) A is false, R is true.
Ans: (A)
Sol: o-nitrophenol forms intramolecular H-bonds, reducing intermolecular association. This makes it more volatile (lower boiling point) compared to p-nitrophenol (which forms intermolecular H-bonds). Hence the mixture can be separated by fractional distillation. Both A and R are true and R correctly explains A.
Q11: Assertion (A): Aquatic species are more comfortable in cold water than in warm water.
Reason (R): Solubility of oxygen gas in water decreases with increase in temperature. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Ans: (A)
Both A and R are true and R is the correct explanation of A. According to Henry's law, the solubility of a gas in a liquid decreases with rise in temperature. Cold water therefore contains more dissolved O2, making it more comfortable for aquatic species.
Very Short Answer Type Questions
Q12: (A) Give reasons: [2 Marks]
(a) Cooking is faster in pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
Ans:
(a) Due to the high pressure maintained inside a pressure cooker, the boiling point of water increases. Because water boils at a higher temperature, the food receives more heat energy at a faster rate, making the cooking process quicker than in an open pan.
(b)
- Type of Deviation: The solution shows Negative Deviation from Raoult's law. This occurs because the new intermolecular forces (X-Y) are stronger than the original forces (X-X and Y-Y), causing the molecules to come closer and reduce the total volume (\(\Delta V_{mix} < 0\)).
- Temperature Change: The temperature of the mixture will increase. Negative deviation is an exothermic process (\(\Delta H_{mix} < 0\)), meaning heat is evolved during mixing.
Given:
Mass of solute (\(w_2\)) = 3 g
Molar mass of solute (\(M_2\)) = 111 g/mol
Mass of solvent (\(w_1\)) = 260 g = 0.26 kg
\(K_b\) for water = 0.52 K kg \(mol^{-1}\)
Since \(CaCl_2\) dissociates completely: \(CaCl_2 \rightarrow Ca^{2+} + 2Cl^-\), so \(i = 3\).
Formula: $$\Delta T_b = i \cdot K_b \cdot m$$ $$\Delta T_b = i \cdot K_b \cdot \frac{w_2}{M_2 \cdot w_1(kg)}$$
Calculation: $$\Delta T_b = 3 \cdot 0.52 \cdot \frac{3}{111 \cdot 0.26}$$ $$\Delta T_b = \frac{4.68}{28.86}$$ $$\Delta T_b \approx 0.162 \text{ K}$$
Result: The elevation of boiling point is 0.162 K.
OR
(b) Liquids ‘X’ and ‘Y’ form an ideal solution. The vapour pressure of pure ‘X’ and pure ‘Y’ are 120 mm Hg and 160 mm Hg respectively. Calculate the vapour pressure of the solution containing equal moles of ‘X’ and ‘Y’. [2 Marks]
Given:
\(P^\circ_X = 120 \text{ mm Hg}\)
\(P^\circ_Y = 160 \text{ mm Hg}\)
Since moles are equal, mole fraction \(\chi_X = \chi_Y = 0.5\).
Formula (Raoult's Law):
$$P_{total} = P^\circ_X \chi_X + P^\circ_Y \chi_Y$$ Calculation:
$$P_{total} = (120 \cdot 0.5) + (160 \cdot 0.5)$$ $$P_{total} = 60 + 80 = 140 \text{ mm Hg}$$ Result: The vapour pressure of the solution is 140 mm Hg.
Q14: State Henry's law. Why are aquatic species more comfortable in cold water as compared to warm water? [2 Marks]
Ans:
- Henry's Law: 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.
- Mathematically: \(p = K_H \cdot x\), where \(p\) is partial pressure, \(K_H\) is Henry's law constant, and \(x\) is mole fraction of gas in solution.
- Aquatic species: The solubility of oxygen \((\text{O}_2)\) increases with decrease in temperature. Cold water thus contains more dissolved oxygen than warm water, making it more comfortable for aquatic species.
Short Answer Type Questions
Q15: An aqueous solution of NaOH was made and its molar mass from the measurement of osmotic pressure at \(27^\circ C\) was found to be \(25 \text{ g mol}^{-1}\). Calculate the percentage dissociation of NaOH in this solution. [Atomic mass: Na = 23 u, O = 16 u, H = 1 u] [3 Marks]
Ans:
Step 1: Calculate the Theoretical Molar Mass of NaOH
Molar mass (\(M_{theoretical}\)) = \(23 (\text{Na}) + 16 (\text{O}) + 1 (\text{H}) = 40 \text{ g mol}^{-1}\)
Step 2: Calculate Van’t Hoff Factor (\(i\))
\(i = \frac{\text{Normal/Theoretical Molar Mass}}{\text{Abnormal/Observed Molar Mass}}\)
\(i = \frac{40}{25} = 1.6\)
Step 3: Calculate Degree of Dissociation (\(\alpha\))
NaOH dissociates as: \(\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-\) (Total ions, \(n = 2\))
For dissociation: \(i = 1 + (n - 1)\alpha\)
\(1.6 = 1 + (2 - 1)\alpha\)
\(0.6 = \alpha\)
Percentage dissociation = \(\alpha \times 100 = 0.6 \times 100 = 60\%\)
Q16: Shweta mixed two liquids A and B of 10 mL each. After mixing, the volume of the solution was found to be 20.2 mL.
(i) Why was there a volume change after mixing the liquids?
(ii) Will there be an increase or decrease of temperature after mixing?
(iii) Give one example for this type of solution. [3 Marks]
Ans:
(i) The total volume (20.2 mL) is greater than the sum of individual volumes (10 + 10 = 20 mL). This indicates positive deviation from Raoult’s Law. This happens because the A-B attractive forces are weaker than A-A and B-B attractive forces.
(ii) There will be a decrease in temperature. Since it is a positive deviation, the mixing process is endothermic (\(\Delta H_{mix} > 0\)), meaning heat is absorbed from the system.
(iii) One example of this type of solution (positive deviation from Raoult’s law) is: Ethanol and benzene mixture.
Q17: (i) How does sprinkling of salt help in clearing the snow covered roads in hilly areas?
(ii) What happens when red blood cells are kept in 0.5% (mass/vol) NaCl solution? Justify your answer.
(iii) Write an application of reverse osmosis. [3 Marks]
Ans:
(i) Sprinkling salt causes depression in the freezing point of water. This allows the snow to melt at the existing lower temperature, effectively clearing the roads.
(ii) The red blood cells will swell and may burst.
Justification: A 0.5% NaCl solution is hypotonic compared to the fluid inside red blood cells (which is equivalent to 0.9% NaCl solution). Water will move into the cells due to osmosis.
(iii) Application: Desalination of seawater to obtain fresh drinking water.
Q18: Vapour pressure of pure water at 298 K is 24.8 mm Hg. Calculate the lowering in vapour pressure of an aqueous solution which freezes at \(-0.3°C\). [3 Marks]
\((K_f \text{ of water} = 1.86 \text{ K kg mol}^{-1})\)
Ans:
Step 1: Find molality using depression in freezing point.
\(\Delta T_f = K_f \times m\)
\(m = \dfrac{\Delta T_f}{K_f} = \dfrac{0.3}{1.86} = 0.16 \text{ mol kg}^{-1}\)
Step 2: Find mole fraction of solute.
\(m = \dfrac{x_2 \times 1000}{M_A} \Rightarrow x_2 = \dfrac{m \times M_A}{1000} = \dfrac{0.16 \times 18}{1000} = 2.88 \times 10^{-3}\)
Step 3: Apply Raoult's Law for relative lowering of vapour pressure.
\(\dfrac{p_1^0 - p_1}{p_1^0} = x_2\)
\(p_1^0 - p_1 = x_2 \times p_1^0 = 2.88 \times 10^{-3} \times 24.8 = \mathbf{0.07 \text{ mm Hg}}\)
Q19: 0.3 g of acetic acid (M = 60 g mol−1) dissolved in 30 g of benzene shows a depression in freezing point equal to \(\Delta T_f = 0.45°C\). Calculate percentage association if it forms a dimer in the solution. (Given: \(K_f\) for benzene = 5.12 K kg mol−1) [3 Marks]
Ans:
\[\Delta T_f = i \cdot K_f \cdot m \implies 0.45 = i \times 5.12 \times \frac{0.3 \times 1000}{60 \times 30}\]
\[i = \frac{0.45 \times 60 \times 30}{5.12 \times 0.3 \times 1000} \approx 0.527\]
For dimerisation (\(n=2\)): \[\alpha = \frac{i-1}{\frac{1}{n}-1} = \frac{0.527-1}{\frac{1}{2}-1} = \frac{-0.473}{-0.5} = 0.946\]
Percentage association = 94.6%
Case-Based Questions
Q20: Read the case given below and answer the questions that follow:
According to the generally accepted definition of an ideal solution, the same or different species of molecules exercise the same intermolecular force. (This statement is equivalent to saying that the activity of the components is equal to their concentration.) In truth, this condition is fulfilled only in exceptional cases for mixtures (optical isomers, mixtures of isotopes of an element, hydrocarbon mixtures). Ideal solutions can be discussed in limited cases only because very dilute solutions with respect to the solvent behave ideally. This viewpoint is further supported by the fact that Raoult's law was found empirically to describe the behavior of the solvent in dilute solutions and the ideal behavior of the solvent can be derived through thermodynamics via a premise. [4 Marks]
Ans:
Example: Phenol and Aniline OR Chloroform and Acetone.
Reason: This deviation occurs because the intermolecular attractive forces between the different components (A-B) are stronger than those between the same components (A-A and B-B). For example, in a mixture of chloroform and acetone, a new hydrogen bond is formed between the chloroform and acetone molecules which was not present in pure components.
(b) (i) State Raoult’s law for a solution containing volatile components.
Ans: For any solution the partial vapour pressure of each volatile component is directly proportional to its mole fraction.
(ii) Raoult’s law is a special case of Henry’s law. Comment.
Ans:
When p0 = KH
p α ꭓ for both.
Raoult’s law is a special case of Henry’s law because when the solute behaves ideally and its concentration approaches unity, Henry’s constant becomes equal to vapour pressure of pure solvent.
Ans:
- The enthalpy of mixing of the pure components in the ideal solution is Zero/∆mixH = 0.
- The Volume of mixing of the pure components in the ideal solution is Zero. /∆mixV = 0
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