Solutions – JEE Advanced Previous Year Questions with Solutions

Solutions JEE Advanced Previous Year Questions with Solutions given here will enable students to get acquainted with the types of questions and their patterns. Furthermore, with these set of questions students can test their preparation level and practice solving problems. With continued practice, students will be able to revise the topics and learn what concepts are important from an exam perspective. All of this will be a crucial step in completing the overall preparation.

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JEE Advanced Previous Year Questions on Solutions

Question 1. The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm^-3, the molarity of urea solution is ________.

(Given data: Molar masses of urea and water are 60 g mol–1 and 18 g mol–1, respectively)

A. 2.98 M

B. 1.98 M

C. 1.06 M

D. 3.98 M

Solution: (2.98)

Mole fraction of urea in aqueous solution = 0.05

0.05 = Moles of urea / Moles of urea + 900 / 18

0.05 (Moles of urea) + 2.5 = Moles of urea

Moles of Urea = 2.63 moles

Mass of urea = 2.63 × 60 = 158 gm

Mass of solution = 158 + 900 = 1058 gm

Density = 1.2 g cm³

Volume of solution = 1058 / 1.2

= 881.67 ml

Molarity = nsolute / V_{solution(innL)} × 1000

= 2.63 / 881.67 × 1000

= 2.98 ≃ 3 M

Question 2. The molar volume of liquid benzene (density = 0.877g/ml^-1) increases by a factor of 2750 as it vaporises at 20°C and that of liquid toluene (density = 0.867g mL^-1) increases by a factor of 7720 at 20°C. A solution of benzene and toluene at 20°C has a vapour pressure of 46.0 torr. Find the mole fraction of benzene in the vapour above the solution.

Solution: (0.735)

(i)Volume = No. of moles × molar mass / density

(ii) PV = nRT or P = nRT/V

Volume of 1 mole of liq. Benzene = 78 / 0.877

The volume of 1 mole of toluene. 92 / 0.867

In the vapour phase,

At 20^oC, for 1 mole of benzene,

Volume = 1 × 78 × 2750 / 0.877 = 244583.80 mL

= 244.58 L

Similarly for 1 mole of toluene,

Volume = 1 × 92 / 0.867 × 7720

= 819192.61 mL = 819.199 L

As we know that, PV = nrt

For benzene, P^o_B = nRT/V = 1 ×0.0821 × 293 / 244.58 atm

= 0.098 atm

For toluene, P^o_T = nRT/V = 1 × 0.0825 × 293 / 819.19 atm = 0.029 atm

P = P^o_B.X_B + P^o_T.X_T

∵ XB + X_T = 1

∴ X_T = 1 – XB

P = P^o_B.X_B + P^o_T (1 – XB)

Total vapour pressure = 46 torr = 46/760 = 0.060 atm

Thus, 0.060 = 0.098 X_B + 0.029 (1 – X_B)

⇒ 0.060 = 0.098 X_B + 0.029 – 0.029 X_B

⇒ 0.031 = 0.069 X_B

∴ X_B = 0.031/0.069 = 0.45 (in liquid phase)

X_B + XT = 1

XT = 1 – 0.45 = 0.55 (in liquid phase)

Also, P^’_B = P^o_B.X_B = PX’_B

So, 0.098 × 0.045 = 0.060 × X^’_B

X^’_B = 0.098 × 0.45 / 0.060 = 0.735 (in gas phase)

Question 3. For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here x_L and x_M represent mole fractions of L and M, respectively, in the solution. The correct statement(s) applicable to this system is (are)

(A) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed from x_L = 0 to x_L = 1

(B) The point Z represents vapour pressure of pure liquid L and Raoult’s law is obeyed when x_L → 1

(C) The point Z represents vapour pressure of pure liquid M and Raoult’s law is obeyed when x_L → 0

(D) Attractive intermolecular interactions between L-L in pure liquid L and M-M in pure liquid M are stronger than those between L-M when mixed in solution.

Solution: (B and D)

From the graph, it is clear that there is a positive deviation in the partial vapour pressure of L in solution to w.r.t pure L. Therefore option D is correct.

When x_L → 1, then Z will have a value equal to P_L^o (vapour pressure of pure L). Therefore option B is also correct.

P_L^o = x_L × P_Total so when x_m = 0 x_L = 1

P_L^o = 1 × P_Total Z in the given graph, so it obeys Raoult`s law

Question 4. Mixture(s) showing positive deviation from Raoult’s law at 35^oC is/are.

A. Carbon tetrachloride + methanol

B. Carbon disulphide + acetone

C. Benzene + toluene

D. Phenol + aniline

Solution: (A and B)

When the intermolecular attraction between two components A and B in the mixture is the same as between A and A, or B and B, it is a case of an ideal solution.

When the intermolecular attraction between A and B in a mixture is smaller than that between A and A, or B and B, the mixture is more vaporised, b.p is lowered. It is a case of positive deviation from Raoult’s law.

When the intermolecular attraction between A and B is higher than that between A and A, or B and B, the mixture is less vaporised, b.p is increased. It is a case of negative deviation.

(A) Methanol molecules CH₃OH are hydrogen-bonded. In a mixture of CCl₄ and CH₃OH, the extent of H – bonding is decreased. The mixture is more vaporised thus, positive deviation from Raoult’s law.

(B) Acetone molecules have higher intermolecular attraction due to dipole-dipole interaction. With CS₂, this interaction is decreased thus, positive deviation.

(C) A mixture of benzene and toluene forms an ideal solution.

(D) Phenol and aniline have higher interaction due to intermolecular H – bonding. Hence, negative deviation.

Question 5. Benzene and naphthalene form an ideal solution at room temperature. For this process,the true statement(s) is (are)

A. ΔG_system is positive

B. ΔS_system is positive

C. ΔS_surroundings = 0

D. ΔH = 0

Solution: (B, C and D)

According to Raoult’s law, for an ideal solution

ΔH = 0, ΔV_mm = 0

From the relation

ΔG = ΔH − TΔS

Since ΔH=0

∴ ΔG = −ve

i.e. less than zero, and ΔS_sys = +ve

i.e. more than zero.

Question 6. Pure water freezes at 273K and 1 bar. The addition of 34.5g of ethanol to 500g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2K kg mol^-1. The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46g mol^-1] Among the following, the option representing the change in the freezing point is.

Solution: (C)

Depression in freezing point of ethanol and water solution:

∆T_f = iK_fm

= 1× 1 × 34.5 / 46 ÷ 500 / 1000 = 3K

So the freezing point of the solution is 270 K

In options b and d, the freezing point shown is 271 K so both are incorrect

Between option a and c, the vapour pressure of the solution is decreasing with temperature after the freezing point in option a so it is also incorrect while option c seems to be the only possible curve for the given conditions.

Question 7. If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong electrolyte) is –0.0558°C, the number of chloride(s) in the coordination sphere of the complex is. [Kf of water = 1.86 K kg mol^-1]

Solution: (1)

ΔT_f = Kf_f × i × m

0.0558 = 1.86 × i × 0.01

i = 3

Given complex behaves as a strong electrolyte

α = 100%

n = 3 (no. of particles given by complex)

∴ complex is [Co(NH₃)₅Cl]Cl₂

no. of Cl− ions in the coordination sphere of the complex = 1

Question 8. MX₂ dissociates into M²⁺ and Xө ion in an aqueous solution, with a degree of dissociation (α) (α) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in absence of ionic dissociation.

Solution: (2)

MX₂ ⇌ M²⁺ + 2X^–

1−α α 2α

Where α is the degree of dissociation.

I = 1 + 2α {α = 0.5}

Therefore, i = 2.

Question 9. On dissolving 0.5 g of non-volatile, non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm of Hg to 640 mm of Hg. The depression of the freezing point of benzene (in K) upon addition of the solute is ___. [Given data: Molar mass and molar freezing point depression of benzene is 78 g mol^-1 and 5.12 K kg mol^-1]

Solution: (1.02)

P_O − P_S / P_S = n_solute / n_solvent

650 − 640 / 640 = 1 × 0.5 × 78 / M × 39

⇒ Msolute = 64gm

△T_f = K_f × m = 5.12 × 0.5 × 1000 / 64 × 39

⇒△T_f = 1.02.

Question 10. 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 solvents.

On adding an equal number of moles of a non-volatile solute S in equal amount (in kg) of these solvents, the elevation of the 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 _________.

Solution: (0.05)

2 = 2 × K_b(x)m

1 = 2K_b(y)m

K_b(x) / K_b(y) = 2

ΔT_b(x) = (1 − α₁ / 2) K_b(x)m 2S ⇌ S2

1−α α / 2

ΔT_b(y) = (1 − α₁ / 2) K_b(y)m i = 1 − α + α / 2

3 = T_b(x) / T_b(y) = (1 − α₁ / 2) ÷ K_b(x) / (1 − 0.7 / 2) K_b(Y) i = 1 − α / 2

3 = (1 − α₁ / 2) × 2 / (1 − 0.7 / 2)

(1 − α₁ / 2) = 3 × 0.65 / 2 = 1.5 × 0.65

α₁ = 0.05

Solutions – Top 10 Most Important and Expected JEE Mains Questions