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 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…

Year	2026	2025	2024	Total
Questions	14	20	11	45

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

For a strong electrolyte like $NaCl$ $NaCl$, the theoretical van 't Hoff factor $itheory$ $i_{theory}$ is 2 ( $Na^+$ $Na^+$ and $Cl^-$ $Cl^-$). 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 $i$ $i$. As the solution becomes infinitely dilute, interionic attractions approach zero, and the observed $i$ $i$ approaches the theoretical value.

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

Higher concentration $\implies$ more ion-pairing $\implies$ lower $i$ $i$. Given concentrations: $A = 0.1 ~M$ $A = 0.1 \mathrm{~M}$ (highest concentration) $B = 0.01 ~M$ $B = 0.01 \mathrm{~M}$ $C = 0.001 ~M$ $C = 0.001 \mathrm{~M}$ (most dilute)

Therefore, the actual $i$ $i$ values follow the reverse order of concentration: $iA (0.1 ~M) < iB (0.01 ~M) < iC (0.001 ~M)$ .

Execution

Salt	Values of i (for different conc. of a Salt)
Salt	0.1 M	0.01 M	0.001 M
NaCl	1.87	1.94	1.97

$i$ $i$ approaches 2 as the solution becomes very dilute.

Pattern Recognition

For strong electrolytes, effective dissociation (and thus $i$ $i$) increases as dilution increases because ions interfere with each other less.

Chapter Mix

Class 12 Chemistry: Solutions

More Solutions Previous-Year Questions

Q59 jee_main_2026_21_jan_morning Elevation of Boiling Point

Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and

PQ₂

. When 1 g of PQ is dissolved in 50 g of solvent ‘A’.

Δ Tb

was 1.176 K while when 1 g of

PQ₂

is dissolved in 50 g of solvent ‘A’,

Δ Tb

was 0.689 K. (

Kb

of ‘A’ =

5 K kg mol⁻¹

). The molar masses of elements P and Q (in

g mol⁻¹

) respectively, are :

A. $70, 110$ $70, 110$
B. $65, 145$ $65, 145$
C. $60, 25$ $60, 25$
D. $25, 60$ $25, 60$

Solution

Related Formula

Δ Tb = Kb \times m

m = Weight of solute (g)Molar mass of solute \times 1000Weight of solvent (g)

Core Logic

For compound PQ:

(Δ Tb)PQ = Kb \cdot m

1.176 = 5 \times 1M₁ \times (1000)/(50)

M₁ = (5 \times 20)/(1.176) = 85.03 g/mol

For compound $PQ₂$ $PQ_2$:

(Δ Tb)PQ₂ = 5 \times 1M₂ \times (1000)/(50) = 0.689

M₂ = (5 \times 20)/(0.689) = 145.13 g/mol

Let molar mass of P & Q be $MP$ $\mathrm{M}_P$ and $MQ$ $\mathrm{M}_Q$ respectively:

MP + MQ = 85.03 --- (1)

MP + 2MQ = 145.13 --- (2)

Subtracting (1) from (2):

MQ = 145.13 - 85.03 = 60.1 \approx 60 g/mol

Substituting back into (1):

MP + 60.1 = 85.03 MP = 24.93 \approx 25 g/mol

Chapter Mix

Class 12 Chemistry: Solutions

Q72 jee_main_2026_21_jan_evening Osmotic Pressure and Isotonic Solutions

The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is

____ g L⁻¹

. (Nearest integer) Given:

R = 0.08 L atm K⁻¹ mol⁻¹

Assume complete dissociation of

NaCl

(Given: Molar mass of Na and Cl are 23 and

35.5 g mol⁻¹

respectively.)

Numerical Answer. Answer: 15 to 15

Solution

Related Formula

$π = i C R T$ $\pi = i C R T$

Core Logic

Given:

$π = 12 atm$ $\pi = 12 \text{ atm}$
$T = 300 K$ $T = 300 \text{ K}$
$i = 2$ $i = 2$ (for complete dissociation of $NaCl$ $\text{NaCl}$)
$R = 0.08 L atm K⁻¹ mol⁻¹$

12 = 2 \times C \times 0.08 \times 300 12 = 48C C = 0.25 mol/L

Step 1: Calculating Strength

Molar mass of $NaCl = 23 + 35.5 = 58.5 g/mol$ . Strength $= 0.25 \times 58.5 = 14.625 g/L \approx 15 g/L$ .

Pattern Recognition

Sees: osmotic pressure calculation for isotonic solutions with electrolyte dissociation. Trap: Forgetting Van 't Hoff factor $i = 2$ $i = 2$ for $NaCl$ $\text{NaCl}$.

Chapter Mix

Class 12 Chemistry: Solutions

Q73 jee_main_2026_21_jan_evening Elevation in Boiling Point and Vapour Pressure Lowering

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass

= 300 g mol⁻¹

) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is

____ \times 10⁻²

. (Nearest integer) [Given:

Kb

of the solvent

= 5.0 K kg mol⁻¹

] Assume the solution to be dilute and no association or dissociation of X takes place in solution.

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

Δ Tb = Kb \cdot m, P° - PₛP° = χsolute

Core Logic

Using boiling point elevation:

Δ Tb = Kb \times m 0.5 = 5.0 \times m m = 0.1 mol/kg

Moles of solvent Y $= (150)/(300) = 0.5 mol$ $= \frac{150}{300} = 0.5 \text{ mol}$. Moles of solute X $= 0.1 \times 0.150 = 0.015 mol$ .

Relative lowering of vapour pressure $= χsolute = (0.015)/(0.015 + 0.5) \approx (0.015)/(0.5) = 0.03 = 3 \times 10⁻²$ .

Step 1: Final Calculation

Thus, the nearest integer value is 3.

Pattern Recognition

Sees: boiling point elevation combined with relative lowering of vapour pressure calculation. Trap: Confusing solvent mass with solution mass when calculating mole fraction.

Chapter Mix

Class 12 Chemistry: Solutions

Q57 jee_main_2026_22_january_morning Henrys Law

Consider a solution of

CO₂(g)

dissolved in water in a closed container. Which one of the following plots correctly represents variation of log (partial pressure of

CO₂

in vapour phase above water) [y-axis] with log (mole fraction of

CO₂

in water) [x-axis] at

25°C

A. $Graph 1$ $\text{Graph 1}$
B. $Graph 2$ $\text{Graph 2}$
C. $Graph 3$ $\text{Graph 3}$
D. $Graph 4$ $\text{Graph 4}$

Solution

Related Formula

P = KH \cdot X

Taking logarithm on both sides:

P = KH + X

Core Logic

From Henry's Law:

P(g) = X(g) + KH

This equation represents a straight line of the form $y = mx + c$ $y = mx + c$, where: $y = P(g)$ $y = \log P(g)$ $x = X(g)$ $x = \log X(g)$ Slope ( $m$ $m$) = +1 y-intercept ( $c$ $c$) = $KH$ $\log K_{H}$ (which is positive).

The plot of $P$ $\log P$ versus $X$ $\log X$ is a straight line with a positive slope and a positive intercept on the y-axis.

Step 1: Conclusion

Graph 3 corresponds to a straight line with a positive slope and a positive y-intercept.

Pattern Recognition

When variables are multiplied ( $P = KH X$ $P = K_{H} X$), their log plot is always a straight line with a slope of +1 and an intercept equal to the log of the constant.

Chapter Mix

Class 12 Chemistry: Solutions

Q70 jee_main_2026_22_january_morning Henrys Law

Given below are two statements: Statement I: The Henry's law constant

KH

is constant with respect to variations in solution's concentration over the range for which the solutions is ideally dilute. Statement II:

KH

does not differ for the same solute in different solvents. In the light of the above statements, choose the correct answer from the options.

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

Solution

Core Logic

Statement I: $KH$ $K_H$ is independent of concentration as long as the solution behaves ideally (is very dilute). This matches the physical definition of Henry's Law constant. (True)

Statement II: Henry's Law constant ( $KH$ $K_H$) depends on the nature of the gas, the nature of the solvent, and the temperature. Therefore, it absolutely differs for the same gas dissolved in different solvents. (False)

Step 1: Final Conclusion

Statement I is true but Statement II is false.

Pattern Recognition

The constants for solubility laws (like $KH$ $K_H$) are fundamentally tied to the intermolecular interactions between the specific solute and the specific solvent.

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

Chapter Mix

Reference Study Guides

More Solutions Previous-Year Questions

Solution

Related Formula

Core Logic

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Related Formula

Core Logic

Step 1: Calculating Strength

Pattern Recognition

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Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

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Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

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Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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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 $i$ $i$