Assume a living cell with 0.9% (omega/omega) of glucose solution (aqueous). This cell is immersed in another solution…

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

Assume a living cell with 0.9%$0.9\%$ (ω/ω)$(\omega/\omega)$ of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will:

Solution & Explanation

Related Formula

Core Logic

Step 1: Concentration Determination and Osmosis Profile

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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Assume a living cell with $0.9%$ $0.9\%$ $(ω/ω)$ $(\omega/\omega)$ of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water. (Consider the data upto first decimal place only) The cell will: