JEE Previous Year Questions- Solutions with - Free MCQ Test Attempt

Molarity is temperature dependent. In the formula of molarity, you have L of solution. Volume is temperature-dependent. It varies with changing temperature. Because of this, molarity changes if the temp changes. 

ΔTb/Kb = ΔTf/Kf
0.186/18.6 = ΔTb/0.512
ΔTb = 0.1x0.512
= 0.0512.
Boiling point = boiling point (of water) + ΔTb
=> 100 + 0.0512
= 100.0512

The correct answer is option A.
Negative deviation means lower vapour pressure, it suggests high boiling point, thus resultant intermolecular force should be stronger than individual one.

The correct answer is option C
If liquids A and B form an ideal solution, the enthalpy of mixing is zero. For an ideal solution ΔVmixing​=0, ΔHmixing​=0. The Gibbs free energy is always negative and becomes more negative as the temperature is increased.

The correct answer is option C
ΔT = Kf​ × molality × (1+α)(HX ⇌ H+   + X−)
Given molality = 0.2;  degree of ionization is 0.3
=1.85×0.2×(1+0.3)
=0.481K
The freezing point of the solution will be 0.81 K.

The correct answer is Option B.
If liquids A and B form an ideal solution, the enthalpy of mixing is zero.
For ideal solution ΔV_mixing = 0, ΔH_mixing = 0. The Gibbs free energy is always negative and becomes more negative as the temperature is increased.

The correct answer is Option A.
Greater the number of solute particles in the solution, higher the elevation in boiling point. Among the given alternatives, Na₂SO₄ will exhibit the highest boiling point since it will dissociate into 3 ions when dissolved in the solution.

The correct answer is option B
Benzene - methanol shows a positive deviation from Raoult's law.
Benzene reduces H -bonding between methanol molecules and thus, evaporation increases or boiling point decreases, i.e., positive deviation from Raoult's law.Thus  in benzene - methanol mixture the intermolecular forces of attractions are weaker than those in pure liquids.

The correct answer is option D
The statement D is false.
Two sucrose solutions of the same molality prepared in different solvents will have different freezing point depression. This is because the molal depression in freezing point constant K_b​ is different for different solvents.
The freezing point depression = ΔT_f​=K_b​×m
Here, m is the molality of sucrose solution.

The correct answer is option D
For Na₂​SO₄ ​: −i = n
Na₂​SO₄ ​⟶ 2Na⁺ + SO₄²⁻​​
                         n=2
i=1−α+nα
i=1−α+2α
i=1+2α

The correct answer is Option B.
 
The molar mass of benzene is 78 g
 
The molar mass of toluene is 92 g
 
n_Ben = 78/78 =1

n_Tol = 92/46 = 0⋅5

X_Ben = 1 / 1+0⋅5

        = 1⋅5

P_Ben = X_BenP⁰_Ben
             = (1/1⋅5) × 75
            =​  50

The correct answer is option D

The correct answer is option C
Greater the number of solute particles in the solution, higher the elevation in boiling point. Among the given alternatives, Na_2​SO₄​ will exhibit the highest boiling point since it will dissociate into 3 ions when dissolved in the solution.
△ = i × k_b​×m
i × m of Na₂​SO₄​ is highest, hence its boiling point will also be highest.
Na₂​SO₄​         i × m=3×0.01=0.03
KNO_3​            i × m=2×0.01=0.02
Urea             i×m=1×0.015=0.015
Glucose       i×m=1×0.015=0.015
 

The correct answer is Option C.

The molar masses of glucose and water are 180 g/mol and 18 g/mol respectively.
 
Number of moles of glucose = 18/80 = 0.1
Number of moles of water = 178.2/18
 =  9.9
Mole fraction of glucose = 0.1/(0.1+9.9)
= 0.01
Relative lowering in vapour pressure is equal to the mole fraction of glucose.
(760−P)/760 = 0.01
760 - P = 7.6
P = 760 −7.6 = 752.4 torr

The correct answer is option B
weight of acetic acid =2.05×60=123g2.05×60=123g
Weight of  solution =1000×1.02=1020g=1000×1.02=1020g
∴ Weight of water=(1020−123)=897g

=2.285 mol kg^-1

The correct answer is Option A.

Total pressure =290 mm
Vapour pressure of propyl alcohol =200 mm
Mole fraction of ethyl alcohol = 0.6
Mol fraction of propyl alcohol =1− 0.6 = 0.4
 
PT = P⁰_A × χ_A + P⁰_B × (1−χ_A)
290 = 200 × 0.4 + P^o_B × 0.6
0.6 × P⁰B = 290 − 80
P⁰B = 350 mm

The correct answer is option D
For isotonic solutions, the osmotic pressure is the same. When solutions do not associate, Ⅱ = CRT and isotonic solutions will have the same concentration.
Let the molar mass of substance be M. It’s molar concentration will be The molar concentration of urea solution is 
According to question,

60 = 210.0 g/mol.
Hence, the molar mass of the substance is 210.0 g/mol.

The correct answer is Option C.

Mass % of H2SO4=29%
Therefore 100g solution contains 29g H2SO4
​
Let the density of solution (in g/ml) is d
Molarity of solution :

Hence the answer is 1.22g/ml.
 

The correct answer is option C
No. of moles of glucose= 
No. of moles of water= 
Mole fraction of glucose= 
Using Raoult's law, as solute is non-volatile,

P^o_A ​=vapour pressure of pure water
P_A ​=vapour pressure of the solution
x_β​=mole fraction of glucose

The vapor pressure of the mixture=17.325 mm Hg.

Let the amount of A in the mixture be χ_A​ and B be the χ_B​.
P_T​= 760 =P_A^o​χ_A​ + P_B^o​χ_B​
760=520χ_A​ + 1000(1−χ_A​)
480χ_A​=240
χ_A​= 1/2​ or 50%.

The correct answer is Option A.
n-Heptane is non-polar and ethanol is polar. Hence the mixture formed by these two liquids is non-ideal. The stronger forces of attraction between n-heptane-n-heptane and ethanol-ethanol are replaced by weaker forces of attraction between n-heptane-ethanol in the solution. Therefore the escaping tendency of the liquid molecules into the vapour phase increases. This will increase the vapour pressure more than expected from Raoult's law. This is a positive deviation. 
Hence, the correct option is B

The correct answer is option B
Let 
P = x_x​P_x⁰ ​+ x_y​P_y0

on solving for P_y⁰​ and P_x⁰​
We get,
P_x⁰​ = 400 mm of Hg
P_y⁰ ​= 600 mm of Hg

The correct answer is Option A.

Here ΔTf = 0−(−2.8^oC)=2.8
∵ΔTf = Kf [W / (M1×W2)]
∵ 2.8 = 1.86 × W / (62×1)
⇒W = 93.33 ≈ 93 g
 

The correct answer is option C
Molarity = 

=2.05 M