Test: JEE Previous Year Questions- Solutions - NEET MCQ

# Test: JEE Previous Year Questions- Solutions - NEET MCQ

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## 26 Questions MCQ Test Chemistry Practice Tests: CUET Preparation - Test: JEE Previous Year Questions- Solutions

Test: JEE Previous Year Questions- Solutions for NEET 2024 is part of Chemistry Practice Tests: CUET Preparation preparation. The Test: JEE Previous Year Questions- Solutions questions and answers have been prepared according to the NEET exam syllabus.The Test: JEE Previous Year Questions- Solutions MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: JEE Previous Year Questions- Solutions below.
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Test: JEE Previous Year Questions- Solutions - Question 1

### Which of the following concentration factor is affected by change in temperature ?  [AIEEE-2002]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 1

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.

Test: JEE Previous Year Questions- Solutions - Question 2

### For an aqueous solution, freezing point is _0.186ºC . Boiling point of the same solution is (Kƒ = 1.86º K mol-1 kg) and Kb = 0.512º K mol-1 kg) [AIEEE-2002]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 2

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

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Test: JEE Previous Year Questions- Solutions - Question 3

### In a mixture of A and B, components show negative deviation when [AIEEE-2002]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 3

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.

Test: JEE Previous Year Questions- Solutions - Question 4

A pressure cooker reduces cooking time for food because _

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 4
Option D is correct To explain this, first of all I would like to define boiling point- the temp. at which vapour pressure of liquid becomes equal to atmospheric pressure, called B. P.
when we close the cooker then pressure inside the coocker increases as compared to normal atmosphereic pressure, then if we start heating then it will not boils at 100 degree centigrade( as we see in normal cases under atmosphereic pressure) becz the pressure inside the cooker is higher. So it need more temp. to boil that's why at higher temp cooking time reduces.
Test: JEE Previous Year Questions- Solutions - Question 5

If liquids A and B form an ideal solution _

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 5

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.

Test: JEE Previous Year Questions- Solutions - Question 6

In a 0.2 molal aqueous solution of a weak acid HX the degree of ionization is 0.3 . Taking kƒ for water as 1.85, the freezing point of the solution will be nearest to _

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 6

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.

Test: JEE Previous Year Questions- Solutions - Question 7

If liquid A and B form ideal solution, than

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 7

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

Test: JEE Previous Year Questions- Solutions - Question 8

Which one of the following aqueous solutions will exhibit highest boiling point ?

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 8

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.

Test: JEE Previous Year Questions- Solutions - Question 9

Osmotic pressure of 40% (wt./vol.) urea solution is 1.64 atm and that of 3.42% (wt./vol.) cane sugar is 2.46 atm. When equal volumes of the above two solutions are mixed, the osmotic pressure of the resulting solution is -

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 9

Test: JEE Previous Year Questions- Solutions - Question 10

Which of the following liquid pairs shows a positive deviation from Raoult's law ?

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 10

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.

Test: JEE Previous Year Questions- Solutions - Question 11

Which one of the following statement is false?

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 11

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 Kb​ is different for different solvents.
The freezing point depression = ΔTf​=Kb​×m
Here, m is the molality of sucrose solution.

Test: JEE Previous Year Questions- Solutions - Question 12

If a is the degree of dissociation of Na2SO4, the vant Hoff's factor (i) used for calculating the molecular mass is–

[AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 12

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

Test: JEE Previous Year Questions- Solutions - Question 13

Benzene and toluene form nearly ideal solutions. At 20ºC, the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial pressure of benzene at 20ºC for a solution containing 78 g of benzene and 46 g of toluene in torr is

[AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 13

The correct answer is Option B.

The molar mass of benzene is 78 g

The molar mass of toluene is 92 g

nBen = 78/78 =1

nTol = 92/46 = 0⋅5

XBen = 1 / 1+0⋅5

= 1⋅5

PBen = XBenP0Ben
= (1/1⋅5) × 75
=​  50

Test: JEE Previous Year Questions- Solutions - Question 14

Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture ?

[AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 14

The correct answer is option D

Test: JEE Previous Year Questions- Solutions - Question 15

Which one of the following aqueous solutions will exhibit highest boiling point?

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 15

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, Na2​SO4​ will exhibit the highest boiling point since it will dissociate into 3 ions when dissolved in the solution.
△ = i × kb​×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
KNO3​            i × m=2×0.01=0.02
Urea             i×m=1×0.015=0.015
Glucose       i×m=1×0.015=0.015

Test: JEE Previous Year Questions- Solutions - Question 16

18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100º C is -

[AIEEE 2006]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 16

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

Test: JEE Previous Year Questions- Solutions - Question 17

Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is -

[AIEEE 2006]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 17

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

Test: JEE Previous Year Questions- Solutions - Question 18

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be -

[AIEEE 2007]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 18

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 = P0A × χA + P0B × (1−χA)
290 = 200 × 0.4 + PoB × 0.6
0.6 × P0B = 290 − 80
P0B = 350 mm

Test: JEE Previous Year Questions- Solutions - Question 19

A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol-1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 gcm-3, molar mass of the substance will be-

[AIEEE 2007]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 19

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.

Test: JEE Previous Year Questions- Solutions - Question 20

The density (in g mL-1) of a 3.60 M sulphuric acid solution that is 29% H2SO4(Molar mass = 98 g mol-1) by mass will be -

[AIEEE 2007]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 20

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 :

Test: JEE Previous Year Questions- Solutions - Question 21

The vapour pressure of water at 20º C is 17.5 mm Hg. If 18g of glucose (C6H12O6) is added to 178.2 g of water at 20° C, the vapour pressure of the resulting solution will be –

[AIEEE 2008]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 21

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,

PoA ​=vapour pressure of pure water
PA ​=vapour pressure of the solution
xβ​=mole fraction of glucose

The vapor pressure of the mixture=17.325 mm Hg.

Test: JEE Previous Year Questions- Solutions - Question 22

At 80º C , the vapour pressure of pure liquid `A' is 520 mm Hg and that of pure liquid `B' is 1000 mm Hg. If a mixture solution of `A' and `B' boils at 80º C and 1 atm pressure, the amount of `A' in the mixture is (1 atm = 760 mm Hg)

[AIEEE 2008]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 22

Let the amount of A in the mixture be χA​ and B be the χB​.
PT​= 760 =PAo​χA​ + PBo​χB
760=520χA​ + 1000(1−χA​)
480χA​=240
χA​= 1/2​ or 50%.

Test: JEE Previous Year Questions- Solutions - Question 23

A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is correct regarding the behaviour of the solution ?

[AIEEE 2009]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 23

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

Test: JEE Previous Year Questions- Solutions - Question 24

Two liquids X and Y form an ideal solution At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively -

[AIEEE 2009]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 24

The correct answer is option B
Let
P = xx​Px0 ​+ xy​Py0

on solving for Py0​ and Px0
We get,
Px0​ = 400 mm of Hg
Py0 ​= 600 mm of Hg

Test: JEE Previous Year Questions- Solutions - Question 25

Kf for water is 1.86 K kg mol-1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to – 2.8°C ?

[AIEEE-2012]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 25

The correct answer is Option A.

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

Test: JEE Previous Year Questions- Solutions - Question 26

The density of a solution prepared by dissolving 120 g of urea (mol. mass = 60 u) in 1000g of water is 1.15 g/mL. The molarity of this solution is :

[AIEEE-2012]

Detailed Solution for Test: JEE Previous Year Questions- Solutions - Question 26

The correct answer is option C
Molarity =

=2.05 M

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