JEE Advanced Previous Year Questions (2018 - 2024): Solutions | Chemistry for JEE Main & Advanced PDF Download

2024

• ΔT_b is the boiling point elevation
• i is the van't Hoff factor
• K_b is the ebullioscopic constant (which is the same for both solutions since they are dissolved in water)
• m is the molality of the solution

We need to compare the boiling point elevations for both solutions in Vessel-1 and Vessel-2.

For Vessel-1, let the boiling point elevation be ΔT_b1, and for Vessel-2, let it be ΔT_b2.

Let's denote the molar mass of solute Y as M_Y g/mol. According to the problem, the molar mass of solute X is 80% of that of Y, i.e.,

M_X = 0.8 ⋅ M_Y
The van't Hoff factor for X is 1.2 times that of Y, i.e.,
i_X = 1.2 ⋅ i_Y

The molality m of each solution can be calculated using the formula:
m = w₂M ⋅ w₁
Thus, the molality of X and Y solutions are:
For solute X:
m_X = w₂M_X ⋅ w₁ = w₂0.8 ⋅ M_Y ⋅ w₁
For solute Y:
m_Y = w₂M_Y ⋅ w₁
Now, substituting the values into the boiling point elevation formula, we get for Vessel-1:
ΔT_b1 = i_X ⋅ K_b ⋅ m_X = (1.2 ⋅ i_Y) ⋅ K_b ⋅ w₂0.8 ⋅ M_Y ⋅ w₁
For Vessel-2:
ΔT_b2 = i_Y ⋅ K_b ⋅ m_Y = i_Y ⋅ K_b ⋅ w₂M_Y ⋅ w₁
To find the ratio of ΔT_b1 to ΔT_b2:
ΔT_b1ΔT_b2 = (1.2 ⋅ i_Y) ⋅ K_b ⋅ w₂0.8 ⋅ M_Y ⋅ w₁i_Y ⋅ K_b ⋅ w₂M_Y ⋅ w₁
= 1.20.8 = 1.5
This means the elevation of the boiling point for the solution in Vessel-1 is 150% of that in Vessel-2. Thus, the elevation of boiling point for solution in Vessel-1 is 150% of the elevation of the boiling point for the solution in Vessel-2.

2023

Q1: 50 mL of 0.2 molal urea solution (density = 1.012 g mL⁻¹ at 300 K) is mixed with 250 mL of a solution containing 0.06 g of urea. Both the solutions were prepared in the same solvent. The osmotic pressure (in Torr) of the resulting solution at 300 K is _______.
[Use: Molar mass of urea = 60 g mol⁻¹ ; gas constant, R = 62 L Torr K⁻¹ mol⁻¹ ;
Assume, Δ_mix H = 0,  Δ_mix  V = 0]     [JEE Advanced 2023 Paper 2]
Ans: 682
Find the weight of urea in the 0.2 molal solution: 
Given a 0.2 molal solution, there are 0.2 moles of urea in 1000 g of solvent. The weight of urea is thus 0.2 mol × 60 g/mol = 12 g.
Find the weight of the solution: 
The total weight of the solution is the weight of the solvent plus the weight of the solute, or
1000 g + 12 g = 1012 g.
Find the volume of the solution: 
Given the density of the solution, we can find its volume by dividing the total weight of the solution by the density

Find the amount of urea in 50 mL of the 0.2 molal solution:
If 1000 mL of the solution contains 0.2 moles of urea, then 50 mL of the solution contains

Find the amount of urea in the 250 mL solution:
The 250 mL solution contains

Find the total concentration of the solution:

After mixing, the total volume of the solution is 50 mL + 250 mL = 300 mL , and the total amount of urea is 0.01 mol + 0.001 mol = 0.011 mol.
So, the concentration of the solution is

Find the osmotic pressure of the solution:
Finally, the osmotic pressure π of the solution can be found using the formula π = CRT , where C is the concentration, R is the gas constant, and T is the temperature. Substituting the given and calculated values,

2022

Q1: An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35^∘C. The salt remains 90 % dissociated in the solution. The vapour pressure of the solution is 59.724 mm of Hg. Vapor pressure of water at 35^∘C is 60.000 mm of Hg . The number of ions present per formula unit of the ionic salt is _________.     [JEE Advanced 2022 Paper 2]
Ans: 5
Vapour pressure of solution (P_A) of = 59.724 mm of Hg
Vapour pressure of pure water (PA^∘) of = 60.000 mm of Hg
Also, 0.1 mol of an ionic solid is dissolved in 1.8 kg of water and salt remains 90 % dissocated in the solution.

So, total number of moles = 0.01 + 0.09a of non-volatile particles.
Now, mass of water = 1.8 kg = 1.8 × 1.8 × 1000 g
Molar mass of water = 18g

Using the colligative property, relative lowering in vapour pressure,

So, the number of ions present per formula unit of the ionic salt is 5.

2021

Q1: The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x^∘C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y × 10^−2∘C.
(Assume: Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), K_b = 0.5 K kg mol⁻¹; Boiling point of pure water as 100^∘C.)
The value of x is ________.    [JEE Advanced 2021 Paper 1]
Ans: 100.1


Boiling point of solution = 100.1^∘C = X

Q2: The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x^∘C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the boiling points of water in the two solutions A and B is y × 10^−2∘C.
(Assume: Densities of the solutions A and B are the same as that of water and the soluble salts dissociate completely.
Use: Molal elevation constant (Ebullioscopic Constant), K_b = 0.5 K kg mol⁻¹; Boiling point of pure water as 100^∘C.)
The value of |y| is ________.        [JEE Advanced 2021 Paper 1]
Ans: 2.5


In final solution total concentration of all ions :

= 0.15 m
= 0.075^∘C
B.P. of solution ' B ' = 100.075^∘C
B.P. of solution 'A′ = 100.1^∘C
|y| = 100.1 − 100.075

2020

Q1: Liquids A and B form ideal solution for all compositions of A and B at 25^∘C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapour pressure of 0.3 and 0.4 bar, respectively. What is the vapour pressure of pure liquid B in bar?   [JEE Advanced 2020 Paper 2]
Ans: 0.2
Using Raoult's law equation for a mixture of volatile liquids.

By solving equation (i) and (ii)

Thus, the vapour pressure of pure liquid B in bar is 0.2.

2019

Q1: On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapor pressure decreases from 650 mmHg to 640 mmHg. The depression of freezing point of benzene (in K) upon addition of the solute is .............
(Given data: Molar mass and the molal freezing point depression constant of benzene are 78 g mol^-1 and 5.12 K kg mol^-1, respectively).     [JEE Advanced 2019 Paper 1]
Ans: 1.02
When 0.5 g of non-volatile solute dissolve into 39 gm of benzene then relative lowering of vapour pressure occurs. Hence, vapour pressure decreases from 650 mmHg to 640 mmHg.
Given, vapour pressure of solvent (p^∘) = 650 mmHg
Vapour pressure of solution (p_s) = 640 mmHg
Weight of non-volatile solute = 0.5 g
Weight of solvent (benzene) = 39 g
From relative lowering of vapour pressure,




∴ Molar mass of solute = 64 g
From molal depression of freezing point,

2018

Q1: Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapor pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions X_A and X_B, respectively, has vapour pressure of 22.5 Torr. The value of z_A / x_B in the new solution is ___________.     [JEE Advanced 2018 Paper 1]
(given that the vapor pressure of pure liquid A is 20 Torr at temperature T)
Ans: 19
We know,

and for equimolar solutions
Given, p_total = 45 torr for equimolar solution and  = 20 torr

Now, for the new solution from the same formula

Hence, the ratio

Q2: The plot given below shows P − T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents.  
 On addition of equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of boiling point of solvent X is three times that of solvent Y. Solute S is known to undergo dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is ___________.      [JEE Advanced 2018 Paper 1]
Ans: 0.05

Dividing Eq. (1) by Eq. (2), we get 

After adding solute S, molality is the same for both solutions.
Dimerisation is a association property. And for dimerisation van't Hoff factor