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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Hard

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 77 questions

1 Hard

1 of 77

Application +4 | -1

When two liquids

A

and

B

are mixed they form an ideal solution. At certain temperature the V.P. of the solution, that contains 3 moles of

A

and 1 mole of

B

was found to be

550 \mathrm{~mm}

\mathrm{Hg}

. The vapour pressure of a solution containing 4 moles of

A

and 1 mole of

B

was found to be 560

\mathrm{mm}

\mathrm{Hg}

. The vapour pressure of pure liquid

A

View answer and explanation

Correct option: D

600 \mathrm{~mm}

Option C (1-phenylpropan-2-ol) is correct because the reaction of 1-phenyl, 2-chloropropane with aqueous KOH results in a substitution reaction, where the chlorine atom is replaced by a hydroxyl group at the same carbon.

2 Hard

2 of 77

Application +4 | -1

0.004 \mathrm{~M}

solution of

\mathrm{Na_2} \mathrm{SO}_4

is isotonic with a

0.010 \mathrm{~M}

solution of glucose at same temperature. The apparent degree of dissociation of

\mathrm{Na}_2 \mathrm{SO}_4

is:

View answer and explanation

Correct option: C

75 \%

Option C is correct because the apparent degree of dissociation is calculated using the van't Hoff factor and the given concentrations.

3 Hard

3 of 77

Analyzing +4 | -1

One gram of a monobasic acid when dissolved in

100 \mathrm{~g}

of water lowers the freezing point by

0.186^{\circ} \mathrm{C}

. Now,

0.25 \mathrm{~g}

of the same acid are dissolved and titrated with

15.1 \mathrm{~mL}

{N} / 10

alkali. The degree of dissociation of the acid is ___

\left[{K_f}\left({H}_2 {O}\right)=1.86\right]

View answer and explanation

Correct option: A

0.67

The correct answer is A (0.67) because the degree of dissociation is calculated using colligative properties and titration data.

4 Hard

4 of 77

Application +4 | -1

How much ethyl alcohol must be added to 1.00 litre of water so that the solution will freeze at

14^{\circ} {F}

\left(K_f\right.

for water

=1.86^{\circ} {C} / {mol}

)

View answer and explanation

Correct option: A

247.31

Option A (247.31) is correct because it is calculated using the colligative property of freezing point depression and the given Kf value.

5 Hard

5 of 77

Application +4 | -1

Mole fraction of the benzene in the vapour phase
is in equilibrium with a solution of benzene

(P^\circ= 200 \text{ Torr})

and toluene

(P^\circ = 150 \text{ Torr})

having 5 moles
of each, is

View answer and explanation

Correct option: A

0.57

Option A (0.57) is correct because it is calculated using Raoult's law and Dalton's law to find the mole fraction of benzene in the vapor phase.

6 Hard

6 of 77

Application +4 | -1

How many grams of

\text{H}_2 \text{SO}_4

are present in

500 \mathrm{~ml}

0.2 \mathrm{~M} \, \text{H}_2 \text{SO}_4

solution?

View answer and explanation

Correct option: A

9.8 \mathrm{~g}

Option A (

9.8 \mathrm{~g}

) is correct because it is the mass of

\text{H}_2 \text{SO}_4

calculated using the given molarity and volume.

7 Hard

7 of 77

Analyzing +4 | -1

Assume a living cell with

0.9 \%(\omega / \omega)

of glucose solution (aqueous). This cell is immersed in another solution having equal mole fraction of glucose and water.

(Consider the data upto first decimal place only)

The cell will:

View answer and explanation

Correct option: D

show no change in volume since solution is

0.9 \%(\omega / \omega)

The correct answer is D (show no change in volume since solution is

0.9 \%(\omega / \omega)

) because the external solution is approximately isotonic with the cell's internal glucose concentration.

8 Hard

8 of 77

Application +4 | -1

At a particular temperature and at a pressure of

1 \mathrm{~atm},

the mass and volume of

\mathrm{O}_3

that can be dissolved in

1 \mathrm{~L}

\mathrm{H}_2 \mathrm{O}

is '

\mathrm{m}

' gram and '

\mathrm{V}

\mathrm{ml},

respectively. Assuming that the solution obeys Henry's law, the mass and volume of

\mathrm{O}_3

dissolved in

2 \mathrm{~L}

of water at the same temperature and at

5 \mathrm{~atm}

will be

View answer and explanation

Correct option: C

10 \mathrm{~m}

gram,

2 \mathrm{~V} \mathrm{~ml}

Option C (10m gram, 2V ml) is correct because the mass of gas dissolved is proportional to pressure and volume of liquid, while the volume of gas dissolved is independent of pressure.

9 Hard

9 of 77

Application +4 | -1

5 \%

solution (mass/vol.) of canc-sugar is isotonic with

0.877 \%

(mass/vol.) of urea solution. Find molar mass of urea, if molar mass of sugar is 342 .

View answer and explanation

Correct option: A

59.99

Option A (59.99) is correct because it is calculated using the isotonicity condition and the molar masses of the substances.

10 Hard

10 of 77

Understanding +4 | -1

Assertion (A) : Water in a beaker cannot be made to boil by placing it in a bath of boiling water.

Reason (R) : Water in the beaker will be heated to

100^{\circ} \mathrm{C}

but will not boil as for boiling it requires latent heat of steam which is not provided by water bath as both attains

100^{\circ} \mathrm{C}

View answer and explanation

Correct option: C

Both Assertion and Reason are correct and Reason is correct explantion of Assertion

Option C is correct because water in a beaker cannot boil in a boiling water bath as it needs latent heat, which the water bath cannot provide once both reach 100°C.

11 Hard

11 of 77

Application +4 | -1

An aqueous solution containing

1 g

of urea boils at

100.25^{\circ} \mathrm{C}

. The aqueous solution containing

3 \mathrm{~g}

of glucose in the same volume will boil at: (Molecular mass of urea and glucose are 60 and 180 respectively)

View answer and explanation

Correct option: C

100.25^{\circ} \mathrm{C}

Option C (

100.25^{\circ} \mathrm{C}

) is correct because the boiling point elevation depends on the number of moles of solute, and urea and glucose have the same effect in this case.

12 Hard

12 of 77

Application +4 | -1

An ideal solution made by mixing two volatile components A and B. If the vapour pressure of solution is 170 mm Hg, then mole fraction of A in vapour phase is

\left( {P_A^0 = 200\,mm\,Hg,\,P_B^0 = 100\,mm\,Hg\,} \right)

View answer and explanation

Correct option: B

0.82

Option B (0.82) is correct because it involves using Raoult's law and Dalton's law to find the mole fraction of A in the vapor phase.

13 Hard

13 of 77

Application +4 | -1

A beaker containing

20 {~g}

sugar in

100 {~g}

water and another containing

10 {~g}

sugar in

100 {~g}

water are placed under a bell-jar and allowed to stand until equilibrium is reached. How much water will be transferred from one beaker to other?

View answer and explanation

Correct option: A

33.3

The correct answer is A (33.3) because at equilibrium, the water will transfer from the less concentrated solution to the more concentrated solution until both solutions have the same vapor pressure.

14 Hard

14 of 77

Application +4 | -1

Vapour pressure of chloroform

\left( {\mathrm{CHC}{\mathrm{l}_3}} \right)

and dichloromethane

\left( {\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{l}_2}} \right)\;\;at\;25^\circ C

are 200 mm Hg and 41.5 mm Hg respectively. Vapour pressure of the solution obtained by mixing 25.5 g of

\left( {\mathrm{CHC}{\mathrm{l}_3}} \right)

and 40 g of

{\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{l}_2}}

at the same temperature will be :
(Molecular mass of

\mathrm{CHC}{\mathrm{l}_3} = 119.5\;\;u

and molecular mass of

\mathrm{C}{\mathrm{H}_2}\mathrm{C}{\mathrm{l}_2} = 85\;u)

View answer and explanation

Correct option: D

90.952 mm Hg

Option D (90.952 mm Hg) is correct because it's calculated using Raoult's Law, considering the mole fractions and vapor pressures of both components.

15 Hard

15 of 77

Application +4 | -1

Mole fraction of solute in benzene is 0.2 then find molality of solute

View answer and explanation

Correct option: A

3.2

Option A (3.2) is correct because the molality is calculated using the mole fraction of the solute and the molar mass of benzene.

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16 Hard

16 of 77

Application +4 | -1

In a Victor Meyer determination of the relative molecular mass of benzene, the heating vessel was maintained at

120^{\circ} \mathrm{C}.

A mass of

0.1528 \mathrm{~g}

of benzene was used and the volume of displaced air collected over water at

15^{\circ} \mathrm{C},

was

48 \mathrm{~cm}^3.

The barometric pressure was

743 \mathrm{~mm} \mathrm{~Hg}.

Calculate the relative molecular mass of benzene. The vapor pressure of water at

15^{\circ} \mathrm{C}

13 \mathrm{~mm} \mathrm{~Hg}.

View answer and explanation

Correct option: B

78.26

Option B (78.26) is correct because it is calculated using the ideal gas law and accounting for the vapor pressure of water.

17 Hard

17 of 77

Analyzing +4 | -1

A non-volatile solute '

\mathrm{X}

' completely dimerizes in water, if the temperature is below

-3.72^{\circ} \mathrm{C}

and the solute completely dissociates as

\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{Z},

if the temperature is above

100.26^{\circ} \mathrm{C}.

In between these two temperatures (including both temperatures), the solute is neither dissociated nor associated. One mole of '

\mathrm{X}

' is dissolved in

1.0 \mathrm{~kg}

water

\left(K_{\mathrm{b}}=0.52 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}, K_{\mathrm{f}}=1.86 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\right)

. Identify the incorrect information related with the solution.

View answer and explanation

Correct option: C

When the solution is cooled to

-7.44^{\circ} \mathrm{C}, 75 \%

of water present initially will separate as ice

Option D is incorrect because the percentage of water separated as ice when cooled to

-7.44^{\circ} \mathrm{C}

is calculated to be

87.5\%

, not

50\%

18 Hard

18 of 77

Remembering +4 | -1

Scuba divers may experience a condition called To avoids this, the tanks used by scuba divers are filled with air diluted with

View answer and explanation

Correct option: D

Bends, Helium

Option D (Bends, Helium) is correct because scuba divers use helium-diluted air to avoid bends caused by nitrogen bubbles in the blood.

19 Hard

19 of 77

Application +4 | -1

Assuming ideal behaviour, what mass of a non-volatile solid of molar mass

128 \mathrm{~g} \mathrm{~mole}^{-1}

would have to be dissolved in

228 \mathrm{~g}

octane

\left(\mathrm{C}_{\mathrm{g}} \mathrm{H}_{18}\right)

to lower the vapour pressure of octane by

20 \%

View answer and explanation

Correct option: A

64 \mathrm{~g}

Option A (

64 \mathrm{~g}

) is correct because it is calculated using Raoult's law, considering the molar masses and the given vapor pressure reduction.

20 Hard

20 of 77

Application +4 | -1

An ideal solution made by mixing two volatile components A and B. If the vapour pressure of solution is 170 mm Hg, then mole fraction of A in vapour phase is

\left( {P_A^0 = 200\,mm\,Hg,\,P_B^0 = 100\,mm\,Hg\,} \right)

View answer and explanation

Correct option: B

0.82

Option B (0.82) is correct because it is calculated using Raoult's law and Dalton's law to determine the mole fraction of A in the vapor phase.

21 Hard

21 of 77

Application +4 | -1

Mole fraction of solute in benzene is 0.2 then find molality of solute

View answer and explanation

Correct option: A

3.2

Option A (3.2) is correct because the molality is calculated using the mole fraction of the solute and the molar mass of benzene.

22 Hard

22 of 77

Analyzing +4 | -1

The order of boiling points of four equimolar aqueous solutions is

C < B < A < D

. The correct order of their freezing points is

View answer and explanation

Correct option: C

D<B>A<C

Option D (

D<B>A<C

) is correct because elevation in boiling point and depression in freezing point are colligative properties, and the order of freezing points is inverse to the order of boiling points.

23 Hard

23 of 77

Application +4 | -1

4.450 \mathrm{~g}

100 per cent sulphuric acid was added to

82.20 \mathrm{~g}

water and the density of the solution was found to be

1.029 \mathrm{~g} / \mathrm{cc}

25^{\circ} \mathrm{C}

and

1 \mathrm{~atm}

pressure. Calculate

(a) the weight percent

\quad

(b) the mole fraction.

View answer and explanation

Correct option: A

(a) 5.14; (b) 0.0098

Option A (+6) is correct because the oxidation number of Cr in CrO5 is determined considering the presence of peroxide linkages.

24 Hard

24 of 77

Application +4 | -1

If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, the change in freezing point of water

\Delta {T_f}

, when 0.01 mole of sodium sulphate is dissolved in 1 kg of water, is

({k_f} = 1.86\;\mathrm{K}\;\mathrm{kg mo}{\mathrm{l}^{ - 1}})

View answer and explanation

Correct option: B

0.0558\mathrm{ }\;K

Option B (

0.0558\mathrm{ }\;K

) is correct because it is calculated using the van't Hoff factor and the freezing point depression formula.

25 Hard

25 of 77

Analyzing +4 | -1

Assertion: There will not be any change in concentration of an ethanol and water mixture containing

85 \%

ethanol by volume on boiling.

Reason: It is because azeotropes boil out in constant composition.

View answer and explanation

Correct option: D

Assertion is wrong, but reason is correct statement.

Option D (Assertion is wrong, but reason is correct statement) is correct because an 85% ethanol-water mixture is not an azeotrope and will change concentration upon boiling, while azeotropes do boil at a constant composition.

26 Hard

26 of 77

Analyzing +4 | -1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): At

10^\circ C,

the density of a 5 M solution of

KCl

[atomic masses of

K\,\,\& \,\,Cl

are 39 & 35.5

g\,\,mo{l^{ - 1}}

respectively] is

'x'\,\,g\,\,m{l^{ - 1}}.

The solution is cooled to

- 21^\circ C.

The molality of the solution will remain unchanged.
Reason (R) : The molality of a solution does not change with temperature as mass remains unaffected with temperature.
In the light of the above statements, choose the correct answer from the options given below.

View answer and explanation

Correct option: A

Both (A) and (R) are true and (R) is the correct explanation of (A).

Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because molality is independent of temperature, and the assertion is true because the solution doesn't freeze at -21°C.

27 Hard

27 of 77

Application +4 | -1

The lowering in vapour pressure caused by the addition of

100 \mathrm{~g}

of sucrose (molecular mass

=342

) to

1000 \mathrm{~g}

of lowering in water, if the vapour pressure of pure water at

25^{\circ} \mathrm{C}

23.8 \mathrm{~mm} \mathrm{~Hg}

, is

View answer and explanation

Correct option: B

0.125 \mathrm{~mm} \mathrm{~Hg}

Option B (0.125 mm Hg) is correct because it is calculated using Raoult's law for vapor pressure lowering.

28 Hard

28 of 77

Application +4 | -1

The distribution coefficient of

{\mathrm{I}_2}

in between

\mathrm{CC}{\mathrm{l}_4}

and

{\mathrm{H}_2}\mathrm{O}

is 85 in favour of

\mathrm{CC}{\mathrm{l}_4}

25^\circ \mathrm{C}

. If solubility of

{\mathrm{I}_2}

{\mathrm{H}_2}\mathrm{O}

25^\circ \mathrm{C}

is 0.33 g

\mathrm{litr}{\mathrm{e}^{ - 1}},\mathrm{the \;solubility\; of}\,{\mathrm{I}_2}

\mathrm{CC}{\mathrm{l}_4}

is ……..g

\mathrm{litr}{\mathrm{e}^{ - 1}}

View answer and explanation

Correct option: A

28.05

Option A (28.05) is correct because the solubility of I2 in CCl4 is calculated using the distribution coefficient and the solubility in H2O.

29 Hard

29 of 77

Analyzing +4 | -1

Liquids 'A' and 'B' form an ideal solution. At

80^{\circ} \mathrm{C}, \mathrm{P}_{\mathrm{A}}^0=0.4 \mathrm{~bar}

and

\mathrm{P}_{\mathrm{B}}^0=0.8 \mathrm{~bar}.

All the vapour above the liquid solution containing equal moles of both the liquids at equilibrium is collected in another empty vessel and condensed. Now, the condesate is heated to

80^{\circ} \mathrm{C}

and all the vapours above the liquid solution at equilibrium is again collected in another empty vessel and condensed. What is the mole-fraction of 'B' is new condensate?

View answer and explanation

Correct option: A

0.8

The correct answer is A (0.8) because after two condensation cycles, the mole fraction of B is enriched due to its higher vapor pressure.

30 Hard

30 of 77

Application +4 | -1

Dry air was passed successively through a solution of

5 \mathrm{~g}

of a solute in

180 \mathrm{~g}

of water and then through pure water. The loss in weight of solution was

2.50 \mathrm{~g}

and that of pure solvent is

0.04 \mathrm{~g}.

The molecular weight of the solute is

View answer and explanation

Correct option: A

31.25

Option A (31.25) is correct because it is calculated using Raoult's Law and the given weight losses of the solution and solvent.

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Key Takeaways

✓ This set contains 77 single-correct-answer MCQs on SOLUTIONS.
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