Given below are two statements : Statement (I): Molal depression constant Kf is given by (MlRTf/Delta 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:

- **Statement I is correct:** Substituting Delta Sfus = Delta 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 for benzene is greater than that of water, making Statement II false. Keep numerical benchmarks for common solvent colligative constants (Kb, Kf for water and benzene) memorized. Benzene has a far lower enthalpy of fusion and a higher freezing point, resulting in a significantly elevated Kf value. Class 12 Chemistry: Solutions

Given below are two statements : Statement (I): Molal depression constant Kf is given by (MlRTf/Delta Sfus), where symbols have their usual meaning. Statement (II): Kf for benzene is less than the Kf

- **Statement I is correct:** Substituting Delta Sfus = Delta 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 for benzene is greater than that of water, making Statement II false. Keep numerical benchmarks for common solvent colligative constants (Kb, Kf for water and benzene) memorized. Benz

Given below are two statements : Statement (I): Molal depression constant Kf is given by (MlRTf/Delta 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:

- **Statement I is correct:** Substituting Delta Sfus = Delta 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 for benzene is greater than that of water, making Statement II false. Keep numerical benchmarks for common solvent colligative constants (Kb, Kf for water and benzene) memorized. Benzene has a far lower enthalpy of fusion and a higher freezing point, resulting in a significantly elevated Kf value. Class 12 Chemistry: Solutions

Given below are two statements : Statement (I): Molal depression constant Kf is given by (MlRTf/Delta Sfus), where symbols have their usual meaning. Statement (II): Kf for benzene is less than the Kf

- **Statement I is correct:** Substituting Delta Sfus = Delta 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 for benzene is greater than that of water, making Statement II false. Keep numerical benchmarks for common solvent colligative constants (Kb, Kf for water and benzene) memorized. Benz

Given below are two statements : Statement (I): Molal depression constant Kf is given by (MlRTf/Delta Sfus), where…

Year	2026	2025	2024	Total
Questions	14	20	11	45

Q54 jee_main_2026_24_january_evening Vapour Pressure of Liquid Solutions

Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and

15kNm⁻²

respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?

A. 0.5217
B. 0.480
C. 0.663
D. 0.340

Solution

Related Formula

YAYB = PA⁰PB⁰ \cdot XAXB

Where $Y$ $Y$ indicates mole fraction in vapour phase and $X$ $X$ indicates mole fraction in liquid phase.

Core Logic

Given data: $PA⁰ = 55 ~kNm⁻²$ $P_A^0 = 55 \mathrm{~kNm}^{-2}$ $PB⁰ = 15 ~kNm⁻²$ $P_B^0 = 15 \mathrm{~kNm}^{-2}$ $YA = 0.8 YB = 1 - 0.8 = 0.2$

Applying the ratio form of Raoult's and Dalton's laws:

(0.8)/(0.2) = (55)/(15) \times XAXB

4 = (11)/(3) \times XAXB

XAXB = (12)/(11)

Step 1: Calculate Mole Fraction in Liquid

Since $XA + XB = 1$ $X_A + X_B = 1$, we can express $XA$ $X_A$ as:

XA = (XA / XB)/(1 + XA / XB)

XA = (12/11)/(1 + 12/11) = (12)/(23)

XA 0.5217

Pattern Recognition

When dealing with equilibrium between liquid and vapour states, taking the ratio $(YA)/(YB) = (PA)/(PB)$ $\frac{Y_A}{Y_B} = \frac{P_A}{P_B}$ bypasses finding the total pressure $Ptotal$ $P_{\text{total}}$ directly and simplifies fraction algebra.

Chapter Mix

Class 12 Chemistry: Solutions

Q68 jee_main_2026_24_january_evening Henry's Law

At 298 K, the mole percentage of

N₂(g)

in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of

N₂(g)

in water at 298 K? (

KH

for

N₂

is

6.5 \times 10⁷

mm Hg)

A. $1.23 \times 10⁻⁷$ $1.23 \times 10^{-7}$
B. $1.17 \times 10⁻⁴$ $1.17 \times 10^{-4}$
C. $9.35 \times 10⁵$ $9.35 \times 10^{5}$
D. $9.35 \times 10⁻⁵$ $9.35 \times 10^{-5}$

Solution

Related Formula

Pgas = KH \cdot Xgas

Core Logic

Given data: Total pressure of air = 10 atm Mole percentage of $N₂$ $N_2$ in air = 80% $PN₂ = (mole fraction in air) × Ptotal = 0.8 × 10 = 8 atm$

Convert partial pressure to mm Hg because $KH$ $K_H$ is given in mm Hg: $PN₂ = 8 \times 760 mm Hg$

Step 1: Apply Henry's Law

PN₂ = KH \cdot XN₂

8 \times 760 = 6.5 \times 10⁷ \times XN₂

XN₂ = (8 \times 760)/(6.5 \times 10⁷)

XN₂ = (6080)/(6.5 \times 10⁷) 935.38 \times 10⁻⁷

XN₂ = 9.35 \times 10⁻⁵

Pattern Recognition

Always align the pressure units. Since $KH$ $K_H$ dictates the unit ecosystem, convert $Pgas$ $P_{\text{gas}}$ to match $KH$ $K_H$ (e.g., atm to mm Hg via $\times 760$ $\times 760$).

Chapter Mix

Class 12 Chemistry: Solutions

Q53 jee_main_2026_28_january_morning Raoults Law for Binary Mixtures

At T(K),

2

moles of liquid A and

3

moles of liquid B are mixed. The vapour pressure of ideal solution formed is

320~mm~Hg

. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as

328.6~mm~Hg

. The vapour pressure (in

mm~Hg

) of A and B are respectively:

A. $300, 200$ $300, 200$
B. $600, 400$ $600, 400$
C. $400, 300$ $400, 300$
D. $500, 200$ $500, 200$

Solution

Related Formula

PS = XA PA^° + XB PB^°

Step 1: First Condition

$2$ $2$ moles of A + $3$ $3$ moles of B (Total $5$ $5$ moles)\n $XA = (2)/(5), XB = (3)/(5)$ \n $320 = PA^° ((2)/(5)) + PB^° ((3)/(5))$ \n $2 PA^° + 3 PB^° = 1600 (I)$

Step 2: Second Condition

Add $1$ $1$ mole of A & $1$ $1$ mole of B (Total $7$ $7$ moles)\n $XA^ = (3)/(7), XB^ = (4)/(7)$ \n $328.6 = PA^° ((3)/(7)) + PB^° ((4)/(7))$ \n $3 PA^° + 4 PB^° = 2300.2 (II)$

Step 3: Solve the Linear Equations

Multiply eq (I) by 3 and eq (II) by 2:\n $6 PA^° + 9 PB^° = 4800$ $$6 P_A^\circ + 9 P_B^\circ = 4800$$\n $6 PA^° + 8 PB^° = 4600.4$ $$6 P_A^\circ + 8 P_B^\circ = 4600.4$$\nSubtracting the two yields:\n $PB^° = 199.6 200~mm~Hg$ \nSubstitute $PB^°$ $P_B^\circ$ back into (I):\n $2 PA^° + 3(200) = 1600 2 PA^° = 1000 PA^° 500~mm~Hg$

Pattern Recognition

Formulating two linear equations based on Raoult's Law mole fractions provides a fast algebraic elimination route to find pure vapour pressures.

Chapter Mix

Class 12 Chemistry: Solutions

Q64 jee_main_2026_28_january_evening Colligative Properties

Consider the following aqueous solutions. I.

2.2 g

Glucose in

125 mL

of solution. II.

1.9 g

Calcium chloride in

250 mL

of solution. III.

9.0 g

Urea in

500 mL

of solution. IV.

20.5 g

Aluminium sulphate in

750 mL

of solution. The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in

g mol⁻¹

:

H=1, C=12, N=14, O=16, Cl=35.5, Ca=40, Al=27 and S=32

]

A. $(1) I < II < III < IV$ $(1)\text{ I < II < III < IV}$
B. $(2) III < I < II < IV$ $(2)\text{ III < I < II < IV}$
C. $(3) II < III < I < IV$ $(3)\text{ II < III < I < IV}$
D. $(4) II < III < IV < I$ $(4)\text{ II < III < IV < I}$

Solution

Related Formula

Δ Tb = i \cdot Kb \cdot m

For dilute solutions, Molarity (M) $\approx$ $\approx$ Molality (m). Thus, $Δ Tb \propto i \times M$ $\Delta T_b \propto i \times M$

Core Logic

Solution	Molarity ( $M = (W)/(Mw) \times (1000)/(V)$ )	Effective Concentration ( $i \times M$ $i \times M$)
(I) Glucose ( $i=1$ $i=1$)	$(2.2)/(180) \times (1000)/(125) = 0.098 M$	$1 \times 0.098 = 0.098$ $1 \times 0.098 = 0.098$
(II) $CaCl₂$ $CaCl_2$ ( $i=3$ $i=3$)	$(1.9)/(111) \times (1000)/(250) = 0.068 M$	$3 \times 0.068 = 0.204$ $3 \times 0.068 = 0.204$
(III) Urea ( $i=1$ $i=1$)	$(9)/(60) \times (1000)/(500) = 0.3 M$	$1 \times 0.3 = 0.300$ $1 \times 0.3 = 0.300$
(IV) $Al₂(SO₄)₃$ $Al_2(SO_4)_3$ ( $i=5$ $i=5$)	$(20.5)/(342) \times (1000)/(750) = 0.080 M$	$5 \times 0.080 = 0.400$ $5 \times 0.080 = 0.400$

Step 1: Compare Effective Concentrations

Comparing the values of $i \times M$ $i \times M$: $0.098 < 0.204 < 0.300 < 0.400$ $0.098 < 0.204 < 0.300 < 0.400$ $\Rightarrow Glucose (I) < CaCl₂ (II) < Urea (III) < Al₂(SO₄)₃ (IV)$

Step 2: Final Conclusion

Order of boiling points corresponds directly to effective concentration. Thus: I < II < III < IV.

Pattern Recognition

For multiple salt mixtures, always multiply molarity by the Van't Hoff factor ( $i$ $i$). Do not compare just raw mass or raw molarity.

Chapter Mix

Class 12 Chemistry: Solutions

Q37 jee_main_2025_02_april_evening Molarity and Temperature Dependency

'x' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from

1°C

to

25°C

. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature? [Consider the solubility of NaCl remains unchanged over the temperature range]

A. $Plot (1)$ $\text{Plot (1)}$
B. $Plot (2)$ $\text{Plot (2)}$
C. $Plot (3)$ $\text{Plot (3)}$
D. $Plot (4)$ $\text{Plot (4)}$

Solution

Related Formula

Molarity (M) = nsoluteVsolution (L)

Core Logic

Since solubility of $NaCl$ $\mathrm{NaCl}$ remains unchanged, the number of dissolved moles of $NaCl$ $\mathrm{NaCl}$ solute ( $nsolute$ $n_{\text{solute}}$) remains strictly constant. Thus, molarity $M$ $M$ is strictly dependent on the volume of water (solvent) as temperature changes:

M \propto 1Vsolution

Step 1: Understand Water's Anomalous Expansion

Water exhibits unique anomalous density behavior near freezing:

From $1°C$ $1^{\circ}\mathrm{C}$ to $4°C$ $4^{\circ}\mathrm{C}$, the density of water increases to a maximum. This contraction means the volume ( $V$ $V$) of water decreases.
From $4°C$ $4^{\circ}\mathrm{C}$ to $25°C$ $25^{\circ}\mathrm{C}$, the density of water decreases due to standard thermal expansion. Consequently, the volume ( $V$ $V$) increases.

Step 2: Relate Volume to Molarity

Because volume is in the denominator of the molarity equation:

From $1°C$ $1^{\circ}\mathrm{C}$ to $4°C$ $4^{\circ}\mathrm{C}$: Volume decreases $\implies$ Molarity increases.
At $4°C$ $4^{\circ}\mathrm{C}$: Volume is minimum $\implies$ Molarity reaches a maximum.
From $4°C$ $4^{\circ}\mathrm{C}$ to $25°C$ $25^{\circ}\mathrm{C}$: Volume increases $\implies$ Molarity decreases.

This behavior is perfectly represented by Plot (2), which features a distinct peak around $4°C$ $4^{\circ}\mathrm{C}$.

Pattern Recognition

Water is at its densest (and occupies minimum volume) at exactly $3.98^$ $3.98^\circ\mathrm{C}$ ( $4^$ $4^\circ\mathrm{C}$). Any concentration unit based on volume (such as Molarity or Normality) will reach a corresponding maximum at this temperature.

Chapter Mix

Class 12 Chemistry: Solutions

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Solutions Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Mole Fraction in Liquid

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Henry's Law

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: First Condition

Step 2: Second Condition

Step 3: Solve the Linear Equations

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compare Effective Concentrations

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Understand Water's Anomalous Expansion

Step 2: Relate Volume to Molarity

Pattern Recognition

Chapter Mix

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