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Test: JEE Main 35 Year PYQs- Solutions - JEE MCQ


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30 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: JEE Main 35 Year PYQs- Solutions

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Test: JEE Main 35 Year PYQs- Solutions - Question 1

Freezing point of an aqueous solution is (–0.186)°C. Elevationof boiling point of the same solution is Kb = 0.512°C, Kf = 1.86°C, find the increase in boiling point.        [2002]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 1

Test: JEE Main 35 Year PYQs- Solutions - Question 2

In mixture A and B components show -ve deviation as                [2002]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 2

In solution containing A and B component showing negative deviation A–A and B–B interactions are weaker than that of A–B interactions. For such solutions.
ΔH = –ve and ΔV = –ve

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Test: JEE Main 35 Year PYQs- Solutions - Question 3

If liquids A and B form an ideal solution           [2003]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 3

When A and B form an ideal solution, ΔHmix = 0

Test: JEE Main 35 Year PYQs- Solutions - Question 4

In a 0.2 molal aqueous solution of a weak acid HX the degreeof ionization is 0.3. Taking kf for water as 1.85, the freezingpoint of the solution will be nearest to [2003]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 4

ΔTf = Kf × m × i ;
ΔTf = 1.85 × 0.2 × 1.3 = 0.480º C
∴Tf= 0 – 0.480ºC = – 0.480ºC

Test: JEE Main 35 Year PYQs- Solutions - Question 5

A pressure cooker reduces cooking time for food because[2003]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 5

NOTE : On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the b.p. of water increases

Test: JEE Main 35 Year PYQs- Solutions - Question 6

Which one of the following aqueous solutions will exihibithighest boiling point ? [2004]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 6

NOTE : Elevation in boiling point is a colligative property, which depends upon the no. of particles.Thus greater the number of particles, greater is it elevation and hence greater will be its boiling point.

Since Na2SO4 has maximum number of particles (3) hence has maximum boiling point.

Test: JEE Main 35 Year PYQs- Solutions - Question 7

For which of the following parameters the structural isomersC2H5OH and CH3OCH3 would be expected to have the samevalues?(Assume ideal behaviour) [2004]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 7

Gaseous densities of ethanol and dimethyl ether would be same at same temperature and pressure. The heat of vaporisation, V.P. and b.pts will differ due to H-bonding in ethanol.

Test: JEE Main 35 Year PYQs- Solutions - Question 8

Which of the following liquid pairs shows a positivedeviation from Raoult’s law ? [2004]

Test: JEE Main 35 Year PYQs- Solutions - Question 9

Which one of the following statements is FALSE? [2004]

Test: JEE Main 35 Year PYQs- Solutions - Question 10

Benzene and toluene form nearly ideal solution. At 20°C,the vapour pressure of benzene is 75 torr and that of tolueneis 22 torr. The partial vapour pressure of benzene at 20°C fora solution containing 78 g of benzene and 46 g of toluene intorr is [2005]

Test: JEE Main 35 Year PYQs- Solutions - Question 11

Equimolar solutions in the same solvent have [2005]

Test: JEE Main 35 Year PYQs- Solutions - Question 12

Among the following mixtures, dipole-dipole as the majorinteraction, is present in [2006]

Test: JEE Main 35 Year PYQs- Solutions - Question 13

18 g of glucose (C6H12O6) is added to 178.2 g of water. Thevapour pressure of water for this aqueous solution at 100ºC is [2006]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 13
Molecular mass of water = 2*1 + 1*16 = 18 g
For 178.2 g water, nA = 9.9
Molecular mass of glucose = 6*12 + 12*1 + 6*16 = 180 g
For 18 g glucose, nB = 0.1
XB = 0.1/(0.1+9.9) = 0.01
XA = 0.99
For lowering of vapour pressure,
P = p^0A^XA = p^0A(1 – XB)
P = 760(1 – 0.01)
= 760 - 7.6
= 752.4 torr
The correct option is B.
Test: JEE Main 35 Year PYQs- Solutions - Question 14

A mixture of ethyl alcohol and propyl alcohol has a vapourpressure of 290 mm at 300 K. The vapour pressure of propylalcohol is 200 mm. If the mole fraction of ethyl alcohol is0.6, its vapour pressure (in mm) at the same temperature willbe [2007]

Test: JEE Main 35 Year PYQs- Solutions - Question 15

Equal masses of methane and oxygen are mixed in an emptycontainer at 25°C. The fraction of the total pressure exertedby oxygen is [2007]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 15
Molar mass of methane = 16.042g mol-1 
Molar mass of oxygen = 32.00 g mol-1 
Therefore if say, 32g of methane and 32g of oxygen mixed,
there is 2 moles of methane and 1 moles of oxygen. 
n= PV/RT 
pressure is directly proportional to the number of moles 
Oxygen is 1/3 of total number of moles and hence it exerts 1/3 of the total pressure 
So my answer would be (D)
Test: JEE Main 35 Year PYQs- Solutions - Question 16

A 5.25% solution of a substance is isotonic with a 1.5% solutionof urea (molar mass = 60 g mol–1) in the same solvent. If thedensities of both the solutions are assumed to be equal to 1.0g cm–3, molar mass of the substance will be [2007]

Test: JEE Main 35 Year PYQs- Solutions - Question 17

At 80° C, the vapour pressure of pure liquid ‘A’ is 520 mmHg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixturesolution of ‘A’ and ‘B’ boils at 80° C and 1 atm pressure, theamount of ‘A’ in the mixture is (1 atm = 760 mm Hg) [2008]

Test: JEE Main 35 Year PYQs- Solutions - Question 18

The vapour pressure of water at 20° C is 17.5 mm Hg. If 18 gof glucose (C6H12O6) is added to 178.2 g of water at 20° C,the vapour pressure of the resulting solution will be [2008]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 18
In solution containing nonvolatile solute,pressure  is directly proportional to its mole fraction.
Psolution = vapour pressure of its pure component * mole fraction in solution
Test: JEE Main 35 Year PYQs- Solutions - Question 19

A binary liquid solution is prepared by mixing n-heptaneand ethanol. Which one of the following statements iscorrect regarding the behaviour of the solution? [2009]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 19

For this solution intermolecular interactions between n-heptane and ethanol aare weaker than n-heptane - nheptane & ethanol-ethanol interactions hence the solution of n-heptane and ethanol is non-ideal and shows positive deviation from Raoult’s law

Test: JEE Main 35 Year PYQs- Solutions - Question 20

Two liquids X and Y form an ideal solution. At 300 K, vapourpressure of the solution containing 1 mol of X and 3 mol ofY is 550 mmHg. At the same temperature, if 1 mol of Y isfurther added to this solution, vapour pressure of thesolution increases by 10 mmHg. Vapour pressure ( in mmHg)of X and Y in their pure states will be, respectively: [2009]

Test: JEE Main 35 Year PYQs- Solutions - Question 21

If sodium sulphate is considered to be completelydissociated into cations and anions in aqueous solution,the change in freezing point of water (ΔTf), when 0.01 molof sodium sulphate is dissolved in 1 kg of water, is(Kf = 1.86 K kg mol–1) [2010]

Test: JEE Main 35 Year PYQs- Solutions - Question 22

On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components(heptane and octane) are 105 kPa and 45 kPa respectively.Vapour pressure of the solution obtained by mixing 25.0 gof heptane and 35 g of octane will be(molar mass of heptane = 100 g mol–1 and of octane = 114 gmol–1) [2010]

Detailed Solution for Test: JEE Main 35 Year PYQs- Solutions - Question 22
Using Raoult's law:-Total vapour pressure=(partial vapour pressure of heptane + partial vapour pressure of octane)= P1X1 + P2X2Where X is mole fraction of component
100 g heptane = 1 mole. 25 g heptane = 0.25 moles. 114 g octane = 1 mole35 g octane = 35/114 = 0.307 moles
Mole fraction of heptane = n1/(n1 + n2)=0.25/(0.25+0.307)=0.4488
Mole fraction of octane = n2/(n1 + n2)=0.307/(0.25+0.307)=0.5511
Applying this in Raoult's law105(0.4488)+45(0.5511)You get 72 kPa
Test: JEE Main 35 Year PYQs- Solutions - Question 23

A 5.2 molal aqueous solution of methyl alcohol, CH3OH, issupplied. What is the mole fraction of methyl alcohol in thesolution? [2011]

Test: JEE Main 35 Year PYQs- Solutions - Question 24

Ethylene glycol is used as an antifreeze in a cold climate.Mass of ethylene glycol which should be added to 4 kg ofwater to prevent it from freezing at –6°C will be : (Kf forwater = 1.86 K kg mol–1, and molar mass of ethylene glycol= 62 g mol–1) [2011]

Test: JEE Main 35 Year PYQs- Solutions - Question 25

The degree of dissociation (a) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression [2011]

Test: JEE Main 35 Year PYQs- Solutions - Question 26

The density of a solution prepared by dissolving 120 g ofurea (mol. mass = 60 u) in 1000 g of water is 1.15 g/mL. Themolarity of this solution is : [2012]

Test: JEE Main 35 Year PYQs- Solutions - Question 27

Kf for water is 1.86 K kg mol–1. If your automobile radiatorholds 1.0 kg of water, how many grams of ethylene glycol(C2H6O2) must you add to get the freezing point of thesolution lowered to –2.8ºC ? [2012]

Test: JEE Main 35 Year PYQs- Solutions - Question 28

The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be : [JEE M 2013]

Test: JEE Main 35 Year PYQs- Solutions - Question 29

Consider separate solutions of 0.500 M C2H5OH(aq),0.100 M Mg3(PO4)2 (aq), 0.250 M KBr(aq) and 0.125 MNa3PO4(aq) at 25 C ° . Which statement is true about thesesolutions, assuming all salts to be strong electrolytes?      [JEE M 2014]

Test: JEE Main 35 Year PYQs- Solutions - Question 30

The vapour pressure of acetone at 20°C is 185 torr. When1.2 g of a non-volatile substance was dissolved in 100 g ofacetone at 20°C, its vapour pressure was 183 torr. The molarmass (g mol–1) of the substance is : [JEE M 2015]

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