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JEE • CHEMISTRY • SOLUTIONS

Solutions practice · Mixed difficulty

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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 460 questions

1 Medium

1 of 460

Application +4 | -1

An aqueous solution of a solute which neither associates nor dissociates has a freezing point depression of

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

An equimolal solution of a second salt has a freezing point depression of

3 \mathrm{X}^{\circ} \mathrm{C}.

Assuming

100 \%

dissociation of salt, the second solution could be a salt of formula

View answer and explanation

Correct option: B

\mathrm{AB}_2

Option B (

\mathrm{AB}_2

) is correct because the van't Hoff factor for this salt is 3, matching the given freezing point depression ratio.

2 Medium

2 of 460

Understanding +4 | -1

Read the following statements carefully and choose the correct option

(i) Osmotic pressure is not a colligative property.

(ii) For dilute solutions, osmotic pressure is proportional to the molarity,

{C}

of the solution at a given temperature

T

.

(iii) During osmosis, solvent molecules always flow from higher concentration to lower concentration of solution.

(iv) The osmotic pressure has been found to depend on the concentration of the solution

View answer and explanation

Correct option: B

(ii) and (iv) are correct

Option B is correct because osmotic pressure is a colligative property that depends on the concentration of the solution, making statements (ii) and (iv) correct.

3 Medium

3 of 460

Remembering +4 | -1

A solution that obeys Raoult’s law is called

View answer and explanation

Correct option: C

Ideal solution

Option C (Ideal solution) is correct because an ideal solution is defined as one that obeys Raoult's law over the entire composition range.

4 Hard

4 of 460

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.

5 Easy

5 of 460

Application +4 | -1

Equal masses of two aqueous solutions, one of

3.6 \, \mathrm{g}

glucose per

100 \mathrm{~g}

of water and the other of

3.6 \mathrm{~g}

urea per

100 \mathrm{~g}

of water are mixed. If the freezing point of resulting solution is

-x^{\circ} \mathrm{C},

then the value of '

1000 x

' is

\left(K_{\mathrm{f}}\right.

of water

\left.=1.86\right)

View answer and explanation

Correct option: A

744

Option A (744) is correct because it calculates the freezing point depression of the mixed solution and multiplies by 1000.

6 Medium

6 of 460

Analyzing +4 | -1

Information

I. Semipermeable membrane contains network of submacroscopic holes or pores.

II. Semipermeable membrane appears to be continuous sheets or films.

III. Solvent molecules cannot pass through the holes of semipermeable membrane but solute molecules can pass On the basis of the information given above select the correct option.

View answer and explanation

Correct option: C

Only III

Option C (Only III) is correct because semipermeable membranes allow solvent molecules to pass through but hinder the passage of larger solute molecules.

7 Medium

7 of 460

Analyzing +4 | -1

Assertion (A) : Addition of

\mathrm{HgI}_2

to aqueous solution of

\mathrm{KI}

shows an increase in vapour pressure.

Reason (R) : The number of particles present in solution decreases due to formation of complex

\mathrm{K}_2 \mathrm{HgI}_4

View answer and explanation

Correct option: C

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

Option C (Both Assertion and Reason are correct and Reason is correct explantion of Assertion) is correct because the addition of

\mathrm{HgI}_2

to KI decreases the number of particles due to complex formation, leading to an increase in vapor pressure.

8 Medium

8 of 460

Analyzing +4 | -1

Three solutions have been provided

I.

3 \mathrm{~g} \mathrm{~HCl}

1 \mathrm{~kg} \mathrm{~H}_2 \mathrm{O}

II.

2 \mathrm{~L} \mathrm{~HCl}(g)

and

1 \mathrm{~L} \mathrm{~H}_2 \mathrm{O}

at room temperature

III. Aqueous solution of

\mathrm{HCl}

with

\chi_{\mathrm{HCl}}=1.5 \times 10^{-3}

.

Solutions containing same number of moles of

\mathrm{HCl}

is/are

View answer and explanation

Correct option: B

I and III

Option D (Statement II is incorrect but Statement I is correct) is correct because the equilibrium constant Kc is indeed 1, but doubling the initial moles of A and B does not double the percentage dissociation.

9 Hard

9 of 460

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.

10 Medium

10 of 460

Analyzing +4 | -1

Assertion : Ifred blood cells were removed from the body and placed in pure water, pressure inside the cells increases.

Reason : The concentration of salt content in the cells increases.

View answer and explanation

Correct option: C

If the Assertion is correct but Reason is incorrect.

Option C is correct because if red blood cells are placed in pure water, water enters the cells due to osmosis, increasing the pressure inside. The salt concentration inside decreases, not increases.

11 Medium

11 of 460

Application +4 | -1

Exactly

1 {~g}

of urea dissolved in

75 {~g}

of water gives a solution that boils at

100.114^{\circ} {C}

at 760

torr

. The molecular weight of urea is 60.1 . The boiling point elevation constant for water is

View answer and explanation

Correct option: B

0.51

Option B (

0.51

) is correct because the boiling point elevation constant is calculated using the colligative properties formula.

12 Easy

12 of 460

Application +4 | -1

The values of observed and calculated molecular weights of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is :

View answer and explanation

Correct option: B

83.5\%

Option B (

83.5\%

) is correct because it calculates the degree of dissociation using the observed and calculated molecular weights.

13 Hard

13 of 460

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.

14 Medium

14 of 460

Application +4 | -1

What is the freezing point of one litre of an aqueous solution of a non-electrolyte having an osmotic pressure of 2 atm at

300 \mathrm{~K}

View answer and explanation

Correct option: D

272.849 \mathrm{~K}

Option D (272.849 K) is correct because it is the freezing point calculated using the osmotic pressure to find molality and then applying the freezing point depression formula.

15 Medium

15 of 460

Application +4 | -1

What is the molarity of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

solution if 25 mL is exactly neutralised with

32.63\mathrm{ }\;mL\mathrm{ }\;of\mathrm{ }\;0.164\mathrm{ }\;M\;\mathrm{ }NaOH

View answer and explanation

Correct option: A

0.107\;\mathrm{ }M

Option A (

0.107\;\mathrm{ }M

) is correct because it is the molarity of H2SO4 calculated using the neutralization reaction with NaOH and accounting for the basicity of H2SO4.

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16 Medium

16 of 460

Application +4 | -1

Henry's law constant for the solubility of methane in benzene at

298 \mathrm{~K}

4.27 \times 10^5 \mathrm{~mm} \mathrm{Hg}

. Calculate the solubility of methane in benzene at

298 \mathrm{~K}

under

760 \mathrm{~mm~Hg}

View answer and explanation

Correct option: A

1.78 \times 10^{-3}

Option A (

1.78 \times 10^{-3}

) is correct because it is calculated directly from Henry's Law:

p = K_H \times x

, where x is the solubility.

17 Medium

17 of 460

Application +4 | -1

What is the boiling point of a solution of

1.00 \mathrm{~g}

of naphthalene dissolved in

94.0 \mathrm{~g}

of toluene? The normal boiling point of toluene is

110.75^{\circ} \mathrm{C}

and

K_{{x}, {b}}=32.0 \mathrm{~K}

for toluene.

View answer and explanation

Correct option: B

111.0^{\circ} \mathrm{C}

The correct answer is

111.0^{\circ} \mathrm{C}

. The boiling point elevation is calculated using the ebullioscopic constant and the mole fraction of the solute.

18 Hard

18 of 460

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.

19 Hard

19 of 460

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.

20 Easy

20 of 460

Remembering +4 | -1

The molal elevation/depression constant depends upon :

View answer and explanation

Correct option: A

Nature of solvent

Option A (Nature of solvent) is correct because the molal elevation/depression constant is a colligative property that depends only on the solvent.

21 Medium

21 of 460

Application +4 | -1

\mathrm{KMnO}_4

reacts with oxalic acid as:

\mathrm{MnO}_4^{-}+\mathrm{C}_2 \mathrm{O}_4^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}

Hence, 50 ml of

0.04 \mathrm{~M~KMnO}_4

is acidic medium is chemically equivalent to

View answer and explanation

Correct option: C

50 ml of

0.1 \mathrm{~M~H}_2 \mathrm{C}_2 \mathrm{O}_4

Option B (50 ml of 0.2 M H₂C₂O₄) is correct because the number of milliequivalents of KMnO₄ and H₂C₂O₄ are equal.

22 Hard

22 of 460

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.

23 Medium

23 of 460

Application +4 | -1

Five percent solution of a solute

(X)

is isotonic
with 0.855% solution of sucrose (molecular
weight = 342). What is the molecular mass of
solute

(X)?

View answer and explanation

Correct option: D

2000

Option D (2000) is correct because the molecular mass of solute X is calculated using the isotonic condition, equating concentrations of the two solutions.

24 Easy

24 of 460

Remembering +4 | -1

The temperature at which vapour pressure of a solvent in its liquid and solid phase becomes same is called :

View answer and explanation

Correct option: B

f. p.

Option B (f. p.) is correct because the freezing point is defined as the temperature at which the vapor pressure of the liquid and solid phases of a solvent are equal.

25 Medium

25 of 460

Application +4 | -1

An aqueous solution is

6 \%

methanol,

\mathrm{CH}_3 \mathrm{OH}

, by mass with

d=0.988 \mathrm{~g} \mathrm{~mL}^{-1}

. Thus, molarity of

\mathrm{CH}_3 \mathrm{OH}

in this solution is

View answer and explanation

Correct option: D

1.85 M

Option D (1.85 M) is correct because the molarity is calculated by finding the moles of methanol and dividing by the volume of the solution in liters.

26 Hard

26 of 460

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.

27 Easy

27 of 460

Analyzing +4 | -1

Assertion (A) Osmotic pressure of

0.1\, {~N}

urea solution is less than that of

0.1 \,{M}\,{NaCl}

solution.

Reason (R) Osmotic pressure is not a colligative property.

View answer and explanation

Correct option: C

If Assertion is true but Reason is false.

Option C (If Assertion is true but Reason is false) is correct because osmotic pressure is a colligative property, and the assertion about urea and NaCl is correct.

28 Medium

28 of 460

Application +4 | -1

What is the molality of pure water?

View answer and explanation

Correct option: C

55.5

Option C (55.5) is correct because the molality of pure water is calculated by dividing 1000g by the molar mass of water (18 g/mol).

29 Medium

29 of 460

Application +4 | -1

A closed container contains equal number of oxygen and hydrogen molecules at a pressure of 740 mm. If oxygen is removed from the system then pressure will:

View answer and explanation

Correct option: B

Become half of 740 mm

Option B is correct because removing oxygen from the system halves the total pressure, as the partial pressures of oxygen and hydrogen were initially equal.

30 Easy

30 of 460

Understanding +4 | -1

Osmotic pressure observed when benzoic acid is dissolved in benzene is less than that expected from theoretical considerations. This is because:

View answer and explanation

Correct option: C

Benzoic acid gets associated in benzene

Option C (Benzoic acid gets associated in benzene) is correct because association of benzoic acid reduces the number of particles in solution, thus lowering the observed osmotic pressure.

Revision notes and key takeaways

Review SOLUTIONS in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 460 single-correct-answer MCQs on SOLUTIONS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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