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.

Statement I: KH 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) 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) Statement I is true but Statement II is false. The constants for solubility laws (like KH) are fundamentally tied to the intermolecular interactions between the specific solute and the specific solvent. Class 12 Chemistry: Solutions

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. St

Statement I: KH 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) 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) Statement I is true but Statement II is false. The constants for solubility laws (like KH) are fundamentally tie

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.

Statement I: KH 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) 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) Statement I is true but Statement II is false. The constants for solubility laws (like KH) are fundamentally tied to the intermolecular interactions between the specific solute and the specific solvent. Class 12 Chemistry: Solutions

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. St

Statement I: KH 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) 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) Statement I is true but Statement II is false. The constants for solubility laws (like KH) are fundamentally tie

Given below are two statements: Statement I: The Henry's law constant KH is constant with respect to variations in…

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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

PQ_2

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

Delta T_b

was 1.176 K while when 1 g of

PQ_2

is dissolved in 50 g of solvent ‘A’,

Delta T_b

was 0.689 K. (

K_b

of ‘A’ =

5text K kg mol^-1

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

textg 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

Delta T_b = K_b times m

m = fractextWeight of solute (g)textMolar mass of solute times frac1000textWeight of solvent (g)

### Core Logic For compound PQ:

(Delta mathrmT_b)_mathrmPQ = mathrmK_b cdot m

1.176 = 5 times frac1mathrmM_1 times frac100050

mathrmM_1 = frac5 times 201.176 = 85.03 text g/mol

For compound

PQ_2

(Delta mathrmT_b)_mathrmPQ_2 = 5 times frac1mathrmM_2 times frac100050 = 0.689

mathrmM_2 = frac5 times 200.689 = 145.13 text g/mol

Let molar mass of P & Q be

mathrmM_P

and

mathrmM_Q

respectively:

mathrmM_P + mathrmM_Q = 85.03 quad text--- (1)

mathrmM_P + 2mathrmM_Q = 145.13 quad text--- (2)

Subtracting (1) from (2):

mathrmM_Q = 145.13 - 85.03 = 60.1 approx 60 text g/mol

Substituting back into (1):

mathrmM_P + 60.1 = 85.03 implies mathrmM_P = 24.93 approx 25 text g/mol

### Evaluation Rubric / Model Answer null ### 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

\_\_\_\_ text g L^-1

. (Nearest integer) Given:

R = 0.08 text L atm K^-1 text mol^-1

Assume complete dissociation of

textNaCl

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

35.5 text g mol^-1

respectively.)

Numerical Answer. Answer: 15 to 15

Solution

### Related Formula

pi = i C R T

### Core Logic Given: -

pi = 12 text atm

T = 300 text K

i = 2

(for complete dissociation of

textNaCl

) -

R = 0.08 text L atm K^-1 text mol^-1

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

### Step 1: Calculating Strength Molar mass of

textNaCl = 23 + 35.5 = 58.5 text g/mol

. Strength

= 0.25 times 58.5 = 14.625 text g/L approx 15 text g/L

. ### Pattern Recognition Sees: osmotic pressure calculation for isotonic solutions with electrolyte dissociation. Trap: Forgetting Van 't Hoff factor

i = 2

for

textNaCl

. ### 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 text 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:

K_b

of the solvent

= 5.0 text 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

Delta T_b = K_b cdot m, quad fracP^circ - P_sP^circ = chi_textsolute

### Core Logic Using boiling point elevation:

Delta T_b = K_b times m implies 0.5 = 5.0 times m implies m = 0.1 text mol/kg

Moles of solvent Y

= frac150300 = 0.5 text mol

. Moles of solute X

= 0.1 times 0.150 = 0.015 text mol

. Relative lowering of vapour pressure

= chi_textsolute = frac0.0150.015 + 0.5 approx frac0.0150.5 = 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

CO_2(g)

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

CO_2

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

CO_2

in water) [x-axis] at

25^circC

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

Solution

### Related Formula

P = K_H cdot X

Taking logarithm on both sides:

log P = log K_H + log X

### Core Logic From Henry's Law:

log P(g) = log X(g) + log K_H

This equation represents a straight line of the form

y = mx + c

, where:

y = log P(g)

x = log X(g)

Slope (

m

) = +1 y-intercept (

c

) =

log K_H

(which is positive). The plot of

log P

versus

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 = 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

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