We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentration 0.1 ~M, 0.01 ~M & 0.001 ~M, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the order.

For a strong electrolyte like NaCl, the theoretical van 't Hoff factor itheory is 2 (Na^+ and Cl^-). However, in real solutions, ion-pairing (interionic attraction) occurs. At higher concentrations, the ions are closer together, leading to stronger interionic attractions that reduce the effective number of independent particles, thus lowering the observed i. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed i approaches the theoretical value. Higher concentration more ion-pairing lower i. Given concentrations: A = 0.1 ~M (highest concentration) B = 0.01 ~M C = 0.001 ~M (most dilute) Therefore, the actual i values follow the reverse order of concentration: iA (0.1 ~M) Salt Values of i (for different conc. of a Salt) 0.1 M 0.01 M 0.001 M NaCl 1.87 1.94 1.97 i approaches 2 as the solution becomes very dilute. For strong electrolytes, effective dissociation (and thus i$) increases as dilution increases because ions interfere with each other less. Class 12 Chemistry: Solutions

We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentration 0.1 ~M, 0.01 ~M & 0.001 ~M, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the

For a strong electrolyte like NaCl, the theoretical van 't Hoff factor itheory is 2 (Na^+ and Cl^-). However, in real solutions, ion-pairing (interionic attraction) occurs. At higher concentrations, the ions are closer together, leading to stronger interionic attractions that reduce the effective number of independent particles, thus lowering the observed i. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed i approaches the theoretical value. Highe

We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentration 0.1 ~M, 0.01 ~M & 0.001 ~M, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the order.

For a strong electrolyte like NaCl, the theoretical van 't Hoff factor itheory is 2 (Na^+ and Cl^-). However, in real solutions, ion-pairing (interionic attraction) occurs. At higher concentrations, the ions are closer together, leading to stronger interionic attractions that reduce the effective number of independent particles, thus lowering the observed i. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed i approaches the theoretical value. Higher concentration more ion-pairing lower i. Given concentrations: A = 0.1 ~M (highest concentration) B = 0.01 ~M C = 0.001 ~M (most dilute) Therefore, the actual i values follow the reverse order of concentration: iA (0.1 ~M) Salt Values of i (for different conc. of a Salt) 0.1 M 0.01 M 0.001 M NaCl 1.87 1.94 1.97 i approaches 2 as the solution becomes very dilute. For strong electrolytes, effective dissociation (and thus i$) increases as dilution increases because ions interfere with each other less. Class 12 Chemistry: Solutions

We have three aqueous solutions of NaCl labelled as 'A', 'B' and 'C' with concentration 0.1 ~M, 0.01 ~M & 0.001 ~M, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the

For a strong electrolyte like NaCl, the theoretical van 't Hoff factor itheory is 2 (Na^+ and Cl^-). However, in real solutions, ion-pairing (interionic attraction) occurs. At higher concentrations, the ions are closer together, leading to stronger interionic attractions that reduce the effective number of independent particles, thus lowering the observed i. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed i approaches the theoretical value. Highe

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

Year	2026	2025	2024	Total
Questions	14	20	11	45

Salt	Values of i (for different conc. of a Salt)
NaCl	1.87	1.94	1.97

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

Q32 jee_main_2025_29_jan_evening Depression of Freezing Point

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:

A. Statement I is false but Statement II is true
B. Both Statement I and Statement II are true
C. Both Statement I and Statement II are false
D. Statement I is true but Statement II is false

Solution

Related Formula

Delta Tf = i cdot Kf cdot m

Core Logic

Statement I is true: Adding NaCl to ice creates a freezing mixture with temperatures below $0circC$ $0^{circ}C$, 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.

Pattern Recognition

This is a classic real-world application of colligative properties. Freezing point lowering keeps commercial refrigeration setups colder for a longer duration.

Chapter Mix

Class 12 Chemistry: Solutions

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

Q41 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. $379.2~K$ $379.2\mathrm{~K}$
B. $377.3~K$ $377.3\mathrm{~K}$
C. $375.3~K$ $375.3\mathrm{~K}$
D. $277.3~K$ $277.3\mathrm{~K}$

Solution

Related Formula

Δ Tb = Kb \cdot mtotal = Kb \cdot ( n₁ + n₂wsolvent(kg))

Tb = Tb^° + Δ Tb

Core Logic

Both ethylene glycol and glucose are non-electrolytes ( $i = 1$ $i = 1$). Total moles of solute $ntotal = 2 + 2 = 4~mol$ . Mass of solvent water $= 500~g = 0.5~kg$ $= 500\mathrm{~g} = 0.5\mathrm{~kg}$.

Step 1: Calculation

Total molality:

mtotal = (4)/(0.5) = 8~mol/kg

Boiling point elevation:

Δ Tb = 0.52 \times 8 = 4.16~K

Boiling point of solution:

Tb = 373.16 + 4.16 = 377.32~K \approx 377.3~K

Pattern Recognition

Colligative properties depend on total moles of non-volatile particles. $Δ Tb = Kb Σ mᵢ = 0.52 \times 8 = 4.16 K Tb = 377.3 K$ .

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.

Chapter Mix

Class 12 Chemistry: Solutions

We have three aqueous solutions of NaCl$NaCl$ labelled as 'A', 'B' and 'C' with concentration 0.1 ~M$0.1 \mathrm{~M}$, 0.01 ~M$0.01 \mathrm{~M}$ & 0.001 ~M$0.001 \mathrm{~M}$, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the order.

Solution & Explanation

Core Logic

Step 1: Correlate Concentration with i$i$

Execution

Pattern Recognition

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Solution

Related Formula

Step 1: Compute Molality

Step 2: Calculate Kf$K_f$

Pattern Recognition

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Related Formula

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Pattern Recognition

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We have three aqueous solutions of $NaCl$ $NaCl$ labelled as 'A', 'B' and 'C' with concentration $0.1 ~M$ $0.1 \mathrm{~M}$, $0.01 ~M$ $0.01 \mathrm{~M}$ & $0.001 ~M$ $0.001 \mathrm{~M}$, respectively. The value of van t’ Hoff factor (i) for these solutions will be in the order.

Step 1: Correlate Concentration with $i$ $i$

Step 2: Calculate $Kf$ $K_f$