XY is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration c1 and c2 (c1 >…

Q: XY is the membrane / partition between two chambers 1 and 2 containing sugar solutions of concentration c1 and c2 (c1 > c2) mol~L-1. For the reverse osmosis to take place identify the correct conditio

pi = c R T where pi is the osmotic pressure of the solution. Given that c1 > c2, chamber 1 has a higher concentration of solute than chamber 2. Under normal conditions, solvent molecules spontaneously flow from lower concentration (chamber 2) to higher concentration (chamber 1) via osmosis. To achieve **reverse osmosis**, the solvent must flow in the opposite direction—from chamber 1 to chamber 2. This requires applying an external pressure on the higher concentration side (chamber 1) that excee

Year	2026	2025	2024	Total
Questions	14	20	11	45

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

$π = c R T$ $\pi = c R T$

where $π$ $\pi$ is the osmotic pressure of the solution.

Core Logic

Given that $c₁ > c₂$ $c_1 > c_2$, chamber 1 has a higher concentration of solute than chamber 2. Under normal conditions, solvent molecules spontaneously flow from lower concentration (chamber 2) to higher concentration (chamber 1) via osmosis.

To achieve reverse osmosis, the solvent must flow in the opposite direction—from chamber 1 to chamber 2. This requires applying an external pressure on the higher concentration side (chamber 1) that exceeds its osmotic pressure $π$ $\pi$.

Condition for Reverse Osmosis: p₁ > π

Cellophane and parchment paper both act as suitable semi-permeable membranes for this setup. Thus, statements (A) and (C) are correct.

Pattern Recognition

Reverse osmosis always requires external pressure applied on the concentrated solution side ( $chigh$ $c_{\text{high}}$) such that $Papplied > π$ $P_{\text{applied}} > \pi$.

Chapter Mix

Class 12 Chemistry: Solutions

Reverse osmosis pressure distribution diagram for Q26 — The diagram illustrates two chambers separated by a membrane XY containing sugar solutions of concentrations c1 and c2.

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

Solution & Explanation

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Solutions Previous-Year Questions

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculating Strength

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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