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…

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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

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

At 298 K, the mole percentage of N₂(g)$N_{2}(g)$ in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of N₂(g)$N_{2}(g)$ in water at 298 K? ( KH$K_{H}$ for N₂$N_{2}$ is 6.5 × 10⁷$6.5 \times 10^{7}$ mm Hg)

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Henry's Law

Pattern Recognition

Chapter Mix

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At 298 K, the mole percentage of $N₂(g)$ $N_{2}(g)$ in air is 80%. Water is in equilibrium with air at a pressure of 10 atm. What is the mole fraction of $N₂(g)$ $N_{2}(g)$ in water at 298 K? ( $KH$ $K_{H}$ for $N₂$ $N_{2}$ is $6.5 \times 10⁷$ $6.5 \times 10^{7}$ mm Hg)