Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circC, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below:

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer duration. Class 12 Chemistry: Solutions

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circ

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer durat

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circC, there is a depression in freezing point. In the light of the above statements, choose the correct answer from the options given below:

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer duration. Class 12 Chemistry: Solutions

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to prevent the melting of ice cream. Statement (II): On addition of NaCl to ice at 0circ

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below 0circC, preventing the ice cream from melting rapidly. Statement II is true: The addition of a non-volatile solute like NaCl causes a depression in the freezing point of water, enabling ice to remain in the solid state at lower surrounding temperatures. This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer durat

Given below are two statements: Statement (I): NaCl is added to the ice at 0circC, present in the ice cream box to…

Year	2026	2025	2024	Total
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More Solutions Previous-Year Questions — Page 5

Q34 jee_main_2025_08_april_evening Colligative Properties

For the weak acid dissociation equation:

HA(aq) leftharpoons H^+(aq) + A^-(aq)

The freezing point depression of a

0.1 m

aqueous solution of this monobasic weak acid HA is found to be

0.20^° C

. The dissociation constant (

Kₐ

) for the acid is: Given:

Kf(H₂O) = 1.8 K kg mol⁻¹

, and assume molality

\approx

molarity.

A. $1.38 \times 10⁻³$ $1.38 \times 10^{-3}$
B. $1.1 \times 10⁻²$ $1.1 \times 10^{-2}$
C. $1.90 \times 10⁻³$ $1.90 \times 10^{-3}$
D. $1.89 \times 10⁻¹$ $1.89 \times 10^{-1}$

Solution

Related Formula

Depression in freezing point colligative relation:

Δ Tf = i \cdot Kf \cdot m

Van 't Hoff factor for a weak acid dissociation: $i = 1 + α$ $i = 1 + \alpha$

Dissociation constant formula:

Kₐ = (Cα²)/(1-α)

Execution

Step 1: Calculate the Van 't Hoff factor $i$ $i$ using experimental freezing data:

0.20 = i \times 1.8 \times 0.1

i = (0.20)/(0.18) = (20)/(18) = (10)/(9)

Step 2: Solve for degree of dissociation $α$ $\alpha$:

i = 1 + α (10)/(9) = 1 + α

α = (10)/(9) - 1 = (1)/(9)

Step 3: Compute $Kₐ$ $K_a$ substituting concentration $C = 0.1 M$ $C = 0.1 \text{ M}$ and $α = (1)/(9)$ $\alpha = \frac{1}{9}$:

Kₐ = (0.1 \times ((1)/(9))²)/(1 - (1)/(9)) = (0.1 \times (1)/(81))/((8)/(9)) = (0.1)/(81) \times (9)/(8) = (0.1)/(72) = (1)/(720)

Kₐ \approx 1.388 \times 10⁻³

Pattern Recognition

When dealing with weak acids, always determine $i$ $i$ first via colligative data, isolate $α$ $\alpha$, and map directly to $Kₐ = (Cα²)/(1-α)$ $K_a = \frac{C\alpha^2}{1-\alpha}$. Speed up computation by converting decimals into fractional fractions ( $0.2/0.18 = 10/9$ $0.2/0.18 = 10/9$) to maintain clean, mistake-free algebra.

Chapter Mix

Class 12 Chemistry: Solutions Class 11 Chemistry: Ionic Equilibrium

Q40 jee_main_2025_28_jan_morning Colligative Properties - Freezing Point Depression

What is the freezing point depression constant of a solvent,

50g

of which contain

1g

non-volatile solute (molar mass

256g mol⁻¹

) and the decrease in freezing point is

0.40K

A. $5.12 K kg mol⁻¹$ $5.12\mathrm{\,K\,kg\,mol}^{-1}$
B. $4.43 K kg mol⁻¹$ $4.43\mathrm{\,K\,kg\,mol}^{-1}$
C. $1.86~K~kg~mol⁻¹$ $1.86\mathrm{~K~kg~mol}^{-1}$
D. $3.72 K kg mol⁻¹$ $3.72\mathrm{\,K\,kg\,mol}^{-1}$

Solution

Related Formula

Freezing point depression relationship:

Δ Tf = Kf \cdot m

where $m$ $m$ is the molality defined as:

m = moles of solutemass of solvent in kg

Step 1: Compute Molality

Moles of non-volatile solute:

moles = 1 g256 g mol⁻¹

Mass of solvent in kg:

mass = 50 g = 50 \times 10⁻³ kg

Therefore, molality values map to:

m = 1256 \times 50 \times 10⁻³ = (1000)/(12800) = (5)/(64) mol kg⁻¹

Step 2: Calculate $Kf$ $K_f$

Substituting values into the core formula:

0.40 = Kf \cdot ((5)/(64))

Kf = (0.40 \times 64)/(5) = 0.08 \times 64 = 5.12 K kg mol⁻¹

Pattern Recognition

Sees: Direct calculation of cryogenic context constant ( $Kf$ $K_f$). Shortcut: Isolate $Kf = Δ Tf \cdot M \cdot Wsolvent1000 \cdot wsolute$ . Substituting instantly returns 5.12.

Chapter Mix

Class 12 Chemistry: Solutions

Q jee_main_2025_03_april_morning Elevation in Boiling Point

2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given: Ebullioscopic constant of water

=0.52 K kg mol⁻¹

)

A. 379.2 K
B. 377.3 K
C. 375.3 K
D. 277.3 K

Solution

Related Formula

The net boiling point elevation for multiple non-volatile solutes is given by:

Δ Tb = (i₁ m₁ + i₂ m₂) Kb

Core Logic

Both ethylene glycol and glucose are non-electrolytes, so their van 't Hoff factors are equal to unity ( $i₁ = i₂ = 1$ $i_1 = i_2 = 1$).

Total moles of solute = 2 + 2 = 4 moles

Mass of solvent (water) = 500 g = 0.5 kg

Total molality (m) = 4 mol0.5 kg = 8 mol/kg

Step 1: Compute Elevation and Final Temperature

Δ Tb = 8 \times 0.52 = 4.16 K

Boiling point of solution = Tb° + Δ Tb = 373.15 K + 4.16 K = 377.31 K \approx 377.3 K

Pattern Recognition

Shortcut: Since both are molecular non-dissociating solutes, simply sum their moles ( $2 + 2 = 4$ $2 + 2 = 4$). Diluting 4 moles in $0.5 kg$ $0.5\text{ kg}$ gives an effective concentration of $8 m$ $8\text{ m}$. Multiplying $8 \times 0.52$ $8 \times 0.52$ gives a shift value of $4.16 K$ $4.16\text{ K}$.

Evaluation Rubric / Model Answer

Option (B)

Chapter Mix

Class 12 Chemistry: Solutions

Q32 jee_main_2025_04_april_evening Colligative Properties

Given below are two statements : Statement (I): Molal depression constant

Kf

is given by

MlRTfΔ Sfus

, where symbols have their usual meaning. Statement (II):

Kf

for benzene is less than the

Kf

for water. In the light of the above statements, choose the most appropriate answer from the options given below:

A. Statement I is incorrect but Statement II is correct
B. Both Statement I and Statement II are incorrect.
C. Both Statement I and Statement II are correct
D. Statement I is correct but Statement II is incorrect

Solution

Related Formula

Kf = M₁ R Tf²Δ Hfus = M₁ R Tf( Δ HfusTf) = M₁ R TfΔ Sfus

Core Logic

Statement I is correct: Substituting $Δ Sfus = Δ HfusTf$ directly matches the given structural relationship formula.
Statement II is incorrect: Standard cryoscopic constants are:
For Benzene: $Kf \approx 5.12 ~^° C \cdot kg \cdot mol⁻¹$
For Water: $Kf \approx 1.86 ~^° C \cdot kg \cdot mol⁻¹$

Therefore, $Kf$ $K_f$ for benzene is greater than that of water, making Statement II false.

Pattern Recognition

Keep numerical benchmarks for common solvent colligative constants ( $Kb, Kf$ $K_b, K_f$ for water and benzene) memorized. Benzene has a far lower enthalpy of fusion and a higher freezing point, resulting in a significantly elevated $Kf$ $K_f$ value.

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Class 12 Chemistry: Solutions

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Solutions Previous-Year Questions — Page 5

Solution

Related Formula

Execution

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Compute Molality

Step 2: Calculate Kf$K_f$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Elevation and Final Temperature

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

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Step 2: Calculate $Kf$ $K_f$