2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
A graph of vapour pressure and temperature for three different liquids X, Y, and Z is shown below :
The following inferences are made :
(A) X has higher intermolecular interactions compared to Y.
(B) X has lower intermolecular interactions compared to Y.
(C) Z has lower intermolecular interactions compared to Y.
The correct inference (s) is / are :
The following inferences are made :
(A) X has higher intermolecular interactions compared to Y.
(B) X has lower intermolecular interactions compared to Y.
(C) Z has lower intermolecular interactions compared to Y.
The correct inference (s) is / are :
A.
(B)
B.
(A)
C.
(C)
D.
(A) and (C)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
Two open beakers one containing a solvent and the other containing a mixture of that solvent
with a non voltatile solute are together selated in a container. Over time :
A.
The volume of the solution increases and the volume of the solvent decreases.
B.
The volume of solution does not change and the volume of the solvent decreases.
C.
The volumer of the solution and the solvent does not change.
D.
The volume of the solution decreases and the volume of the solvent increases.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
At 35oC, the vapour pressure of CS2
is 512 mm. Hg and that of acetone is 344 mm Hg. A solution
of CS2
in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the
following is :
A.
CS2 and acetone are less attracted to each other than to themselves.
B.
a mixture of 100 ml. CS2 and 100 ml. acetone has a volume < 200 ml.
C.
Heat must be absorved in order to produce the solution 35oC
D.
Raoult's law is not obeyed by this system.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar
mass = 180 g mol–1) in 100 mL of water at 27oC. The osmotic pressure of the solution is :
(R = 0.08206 L atm K–1 mol–1)
(R = 0.08206 L atm K–1 mol–1)
A.
8.2 atm
B.
2.46 atm
C.
4.92 atm
D.
1.64 atm
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
1 g of a non-volatile non-electrolyte solute is dissolved in 100 g of two different solvents A and B whose
ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation in their boiling points, ${{\Delta {T_b}(A)} \over {\Delta {T_b}(B)}}$, is :
A.
5 : 1
B.
1 : 0.2
C.
10 : 1
D.
1 : 5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the
vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be.
(molar mass of urea = 60 g mol–1)
A.
0.031 mmHg
B.
0.017 mmHg
C.
0.028 mmHg
D.
0.027 mmHg
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
Molal depression constant for a solvent is
4.0 kg mol–1. The depression in the freezing
point of the solvent for 0.03 mol kg–1 solution
of K2SO4 is :
(Assume complete dissociation of the electrolyte)
(Assume complete dissociation of the electrolyte)
A.
0.18 K
B.
0.24 K
C.
0.36 K
D.
0.12 K
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
Liquid 'M' and liquid 'N' form an ideal solution.
The vapour pressures of pure liquids 'M' and
'N' are 450 and 700 mmHg, respectively, at the
same temperature. Then correct statement is:
(xM = Mole fraction of 'M' in solution ;
xN = Mole fraction of 'N' in solution ;
yM = Mole fraction of 'M' in vapour phase ;
yN = Mole fraction of 'N' in vapour phase)
(xM = Mole fraction of 'M' in solution ;
xN = Mole fraction of 'N' in solution ;
yM = Mole fraction of 'M' in vapour phase ;
yN = Mole fraction of 'N' in vapour phase)
A.
${{{x_M}} \over {{x_N}}} < {{{y_N}} \over {{y_N}}}$
B.
(xM – yM) < (xN – yN)
C.
${{{x_M}} \over {{x_N}}} = {{{y_N}} \over {{y_N}}}$
D.
${{{x_M}} \over {{x_N}}} > {{{y_M}} \over {{y_N}}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
The osmotic pressure of a dilute solution of an
ionic compound XY in water is four times that
of a solution of 0.01 M BaCl2 in water.
Assuming complete dissociation of the given
ionic compounds in water, the concentration of
XY (in mol L–1) in solution is :
A.
4 × 10–4
B.
6 × 10–2
C.
4 × 10–2
D.
16 × 10–4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
For the solution of the gases w, x, y and z in
water at 298K, the Henrys law constants (KH)
are 0.5, 2, 35 and 40 kbar, respectively. The
correct plot for the given data is :-
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298 K on
mixing the two liquids, the sum of their initial volume is equal ot the volume of the final mixture.
The mole fraction of liquid B is 0.5 in the mixture, The vapour pressure of the final solution, the
mole fractions of components A and B in vapour phase, respectively are :
A.
500 mmHg. 0.5,0.5
B.
500 mmHg, 0.4, 0.6
C.
450 mmHg, 0.4,0.6
D.
450 mmHg.0.5,0.5
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
Molecules of benzoic acid (C6H5COOH) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1)
A.
1.5 g
B.
1.8 g
C.
1.0 g
D.
2.4 g
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is -
A.
4A
B.
2A
C.
3A
D.
A
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
K2Hgl4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is:
A.
1.6
B.
2.2
C.
2.0
D.
1.8
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
The freezing point of a diluted milk sample is found to be –0.2oC, while it should have been –0.5oC for pure milk. How much water has been added to pure milk to make the diluted sample?
A.
1 cup of water to 2 cups of pure milk
B.
2 cups of water to 3 cups of pure milk
C.
3 cups of water to 2 cups of pure milk
D.
1 cup of water to 3 cups of pure milk
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
Elevation in the boiling point for 1 molar solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is
A.
Kb = Kf
B.
Kb = 0.5 Kf
C.
Kb = 1.5 Kf
D.
Kb = 2 Kf
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vaapor pressures of pure A and pure B are 7 $ \times $ 103 Pa and 12 $ \times $ 103 Pa, respectively . The composition of the vapor in equilibriumwith a solution containing 40 mole percent of A at this temperature is :
A.
xA = 0.76; xB = 0.24
B.
xA = 0.28; xB = 0.72
C.
xA = 0.4; xB = 0.6
D.
xA = 0.37; xB = 0.63
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
A solution containing 62 g ethylene glycol in 250 g water is cooled to $-$ 10oC. If Kf for water is 1.86 K kg mol$-$1 , the amount of water (in g) separated as ice is :
A.
48
B.
32
C.
64
D.
16
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
Which one of the following statements regarding Henry's law is not correct ?
A.
Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids
B.
Different gases have different KH (Henry's law constant) values at the same temperature.
C.
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
D.
The value of KH increases with increase of temperature and KH is function of the nature of the gas
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol-1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%, is :
A.
37.5 g
B.
75 g
C.
150 g
D.
50 g
2018
JEE Mains
MCQ
JEE Main 2018 (Offline)
For 1 molal aqueous solution of the following compounds, which one will show the highest freezing
point?
A.
[Co(H2O)3 Cl3].3H2O
B.
[Co(H2O)6] Cl3
C.
[Co(H2O)5 Cl] Cl2.H2O
D.
[Co(H2O)4 Cl2] Cl.2H2O
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Evening Slot
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volatile solute separately in the solvents X and Y. The molecular weights of the solvents are Mx and My, respectively where Mx = ${3 \over 4}$ My. The relative lowering of vapor pressure of the solution in X is ''m'' times that of the solution in Y. Given that the number of moles of solute is very small in comparison to that of the solvent, the value of ''m'' is :
A.
${4 \over 3}$
B.
${3 \over 4}$
C.
${1 \over 2}$
D.
${1 \over 4}$
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 9th April Morning Slot
A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3 . If vapour pressure of CH2Cl2
and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is : (Molar mass of Cl = 35.5 g mol−1)
A.
0.162
B.
0.675
C.
0.325
D.
0.486
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 8th April Morning Slot
5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be
3.82oC. If Na2SO4 is 81.5% ionised,
the value of x
(Kf for water=1.86oC kg mol−1) is approximately :
(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)
(Kf for water=1.86oC kg mol−1) is approximately :
(molar mass of S = 32 g mol−1 and that of Na = 23 g mol−1)
A.
15 g
B.
25 g
C.
45 g
D.
65 g
2017
JEE Mains
MCQ
JEE Main 2017 (Offline)
The freezing point of benzene decreases by 0.450C when 0.2 g of acetic acid is added to 20g of benzene. If
acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be:
(Kf for benzene = 5.12 K kg mol–1)
A.
80.4 %
B.
74.6 %
C.
94.6 %
D.
64.6 %
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 10th April Morning Slot
An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :
A.
0.33
B.
0.50
C.
0.67
D.
0.80
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 9th April Morning Slot
The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 torr partial pressure is :
A.
0.0075
B.
0.015
C.
0.02
D.
0.005
2016
JEE Mains
MCQ
JEE Main 2016 (Offline)
18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :
A.
76.0
B.
752.4
C.
759.0
D.
7.6
2015
JEE Mains
MCQ
JEE Main 2015 (Offline)
The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:
A.
64
B.
128
C.
488
D.
32
2014
JEE Mains
MCQ
JEE Main 2014 (Offline)
Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and 0.125
M Na3PO4(aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?
A.
0.125 M Na3PO4(aq) has the highest osmotic pressure.
B.
0.500 M C2H5OH(aq) has the highest osmotic pressure
C.
They all have the same osmotic pressure
D.
0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure.
2012
JEE Mains
MCQ
AIEEE 2012
Kf for water is 1.86K 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 –2.8oC ?
A.
72 g
B.
93 g
C.
39 g
D.
27 g
2011
JEE Mains
MCQ
AIEEE 2011
The degree of dissociation ($\alpha$ ) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :
A.
$\alpha = {{i - 1} \over {x + y + 1}}$
B.
$\alpha = {{x + y - 1} \over {i - 1}}$
C.
$\alpha = {{x + y + 1} \over {i - 1}}$
D.
$\alpha = {{i - 1} \over {(x + y - 1)}}$
2011
JEE Mains
MCQ
AIEEE 2011
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be
added to 4 kg of water to prevent it from freezing at −6oC will be :
[Kf for water = 1.86 K kg mol−1 , and molar mass of ethylene glycol = 62 g mol−1 )
A.
204.30 g
B.
400.00 g
C.
304.60 g
D.
804.32 g
2010
JEE Mains
MCQ
AIEEE 2010
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 25.0g of heptane and 35 g of octane will be (molar mass of heptane = 100 g mol–1 and of octane = 114 g mol–1)
A.
72.0 kPa
B.
36.1 kPa
C.
96.2 kPa
D.
144.5 kPa
2010
JEE Mains
MCQ
AIEEE 2010
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous
solution, the change in freezing point of water (∆Tf), when 0.01 mol of sodium sulphate is dissolved
in 1 kg of water, is (Kf = 1.86 K kg mol–1)
A.
0.0372 K
B.
0.0558 K
C.
0.0744 K
D.
0.0186 K
2009
JEE Mains
MCQ
AIEEE 2009
Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol
of X and 3 mol of Y is 550 mm Hg. At the same temperature, if 1 mol of Y is further added to this
solution, vapour pressure of the solution increases by 10 mm Hg. Vapour pressure (in mm Hg) of X
and Y in their pure states will be, respectively :
A.
200 and 300
B.
300 and 400
C.
400 and 600
D.
500 and 600
2009
JEE Mains
MCQ
AIEEE 2009
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 ?
A.
The solution formed is an ideal solution
B.
The solution is non-ideal, showing +ve deviation from Raoult’s law.
C.
The solution is non-ideal, showing –ve deviation from Raoult’s law.
D.
n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s law.
2008
JEE Mains
MCQ
AIEEE 2008
The vapour pressure of water at 20oC is 17.5 mm Hg. If 18 g of glucose (C6H12O6) is added to 178.2 g of water at 20oC, the vapour pressure of the resulting solution will be
A.
17.675 mm Hg
B.
15.750 mm Hg
C.
16.500 mm Hg
D.
17.325 mm Hg
2008
JEE Mains
MCQ
AIEEE 2008
At 80oC, 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 80oC and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)
A.
52 mol percent
B.
34 mol percent
C.
48 mol percent
D.
50 mol percent
2007
JEE Mains
MCQ
AIEEE 2007
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
A.
360
B.
350
C.
300
D.
700
2007
JEE Mains
MCQ
AIEEE 2007
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 g cm−3, molar mass of the substance will be
A.
90.0 g mol−1
B.
115.0 g mol−1
C.
105.0 g mol−1
D.
210.0 g mol−1
2006
JEE Mains
MCQ
AIEEE 2006
18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100oC is
A.
759.00 Torr
B.
7.60 Torr
C.
76.00 Torr
D.
752.40 Torr
2006
JEE Mains
MCQ
AIEEE 2006
Among the following mixtures, dipole-dipole as the major interaction, is present in
A.
benzene and ethanol
B.
acetonitrile and acetone
C.
KCl and water
D.
benzene and carbon tetrachloride
2005
JEE Mains
MCQ
AIEEE 2005
If $\alpha$ is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for
calculating the molecular mass is :
A.
1 + $\alpha$
B.
1 + $2\alpha$
C.
1 - $\alpha$
D.
1 - $2\alpha$
2005
JEE Mains
MCQ
AIEEE 2005
Benzene and toluene form nearly ideal solutions. At 20 oC, the vapour pressure of
benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of
benzene at 20 oC for a solution containing 78 g of benzene and 46 g of toluene in
torr is
A.
50
B.
25
C.
53.5
D.
37.5
2005
JEE Mains
MCQ
AIEEE 2005
Equimolar solutions in the same solvent have
A.
Same boiling point but different freezing point
B.
Same freezing point but different boiling point
C.
Same boiling and same freezing points
D.
Different boiling and different freezing points
2004
JEE Mains
MCQ
AIEEE 2004
Which one of the following aqueous solutions will exhibit highest boiling point?
A.
0.01 M Na2SO4
B.
0.015 M glucose
C.
0.015 M urea
D.
0.01 M KNO3
2004
JEE Mains
MCQ
AIEEE 2004
For which of the following parameters the structural isomers C2H5OH and CH3OCH3 would
be expected to have the same values? (Assume ideal behaviour)
A.
Heat of vaporization
B.
Gaseous densities at the same temperature and pressure
C.
Boiling points
D.
Vapour pressure at the same temperature
2004
JEE Mains
MCQ
AIEEE 2004
Which one of the following statements is false?
A.
Raoult’s law states that the vapour pressure of a components over a solution is
proportional to its mole fraction
B.
Two sucrose solutions of same molality prepared in different solvents will have the same
freezing point depression
C.
The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is
BaCl2 > KCl > CH3COOH > sucrose
D.
The osmotic pressure ($\pi$) = MRT, where M is the molarity of the solution
2004
JEE Mains
MCQ
AIEEE 2004
Which of the following liquid pairs shows a positive deviation from Raoult’s law?
A.
Water – hydrochloric acid
B.
Acetone – chloroform
C.
Water – nitric acid
D.
Benzene – methanol
