WBJEE 2025 Physical Chemistry Practice Questions with Solutions:Attempting as many WBJEE Physical Chemistry sample questions as you can will make you familiar with the frequently asked and most expected questions in the upcoming examination. Physical Chemistry covers a significant section of theWBJEE 2025 Chemistry syllabus PDF. The WBJEE 2025 Physical Chemistry practice questions comprise some important topics, such as Atomic Structure, Redox Reactions, Gaseous States, Thermodynamics etc. With an approximate weightage of 4-7% in theWBJEE 2025 examthe question paper will cover 8-9 questions from this chapter. However, since the topics are already covered in the West Bengal 11th and 12th standard Board exam curriculum, you are advised to focus on solving WBJEE 2025 Physical Chemistry practice questions rather than spending time studying all the topics from scratch. In this article, we will discuss a series of practice questions for WBJEE 2025 Physical Chemistry chapter in detail. We will also provide you with some of the most important WBJEE 2025 Physical Chemistry practice questions with solutions based on the past years' paper review and format.

Also Check -WBJEE 2025 Properties of Matter Practice Questions with Solutions

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List of WBJEE 2025 Physical Chemistry Sub Topics for Exam Preparations

While solving WBJEE Physical Chemistry sample questions you may come across a number of topics and sub topics that are repetitive and asked more often. These are the sub topics that you need to pay attention to since the majority of the questions will be based around these. In this section we have listed the key sub topics from the Physical Chemistry chapter as mentioned in theWBJEE 2025 syllabus PDF.

While practising with WBJEE Physical Chemistry sample questions you should also be aware of the expected marks weightage and the probable number of questions in the upcoming examination. By analyzing the marks weightage you can figure out which topics to prioritize and accordingly solve WBJEE 2025 Physical Chemistry practice questions from those sub topics. If you manage to attempt 5-6 questions from the chapter, you can score at least 10-12 marks. Let’s have a look at the tentative WBJEE 2025 marks weightage for Physical Chemistry below.

WBJEE 2025 Physical Chemistry Practice Questions with Solutions

Practising regularly with WBJEE Physical Chemistry sample questions can make all the difference in your exam preparations. By attempting WBJEE 2025 Physical Chemistry practice questions with solutions, you can get a glimpse of the probable questions, correct asnwer and the step-by-step solution to each question. Check out some of the important WBJEE 2025 Physical Chemistry practice questions picked from the past years' exam papers by experts.

Q1. Arrange the following in order of increasing mass:-

I. 1 mole of N₂

II. 0.5 mole of O₃

III.3.011 x 10²³molecules of O₂

IV. 0.5 gm atom of O₂

a. IV < III < II < I

b. IV < I < III < II

c. III < II < IV < I

d. I < III < II < IV

Ans: a. IV < III < II < I

Q2. 63 g of a compound (Mol. Wt. = 126) was dissolved in 500 g distilled water. The density of the resultant solution as 1.126 g/ml. The molarity of the solution is:

a. 1.25 M

b. 1.0 M

c. 0.75 M

d. 1.1 M

Ans: b. 1.0 M

Solution:

Given

Mass of compound (Solute) = 63g

Mass of solvent = 500g

Molecular weight = 126g

Density = 1.126g/mL

Molarity =?

Mass of solution = Mass of solute + Mass of solvent

= 63 + 500 = 563g

Number of moles = 63/126 = 0.5 moles

Volume= Mass/Density = 563/1.126 mL

Now, using the formula of molarity,

M = number of moles of solute/volume of solution

1 = (0.5 x 1000)/(563/1.126)

1= 1.126 x 1000 x 0.5/563 = 1 Molar or 1.0M

Q3. How much solid oxalic acid (Molecular weight 126) has to be weighed to prepare 100 ml. exactly 0.1 (N) oxalic acid solution in water?

a. 1.26 g

b. 0.126 g

c. 0.63 g

d. 0.063 g

Ans: c. 0.63 g

Solution:

Given, the molecular weight of oxalic acid = 126

Volume = 100 mL

Normality = 0.1 N

Mass =?

We know,

Normality = (Mass/equivalent weight)/Volume

Equivalent weight= Molecular weight/η-factor = 126/2 [η-factor = 2 for oxalic acid]

0.1 = Mass/(63x100/1000) = Mass = 0.1 x 63 x 100/1000

Mass = 0.63 g

Q4. Consider an electron moving in the first Bohr orbit of a Hw⁺ion with a velocity v₁. If it is allowed to move in the third Bohr orbit with a velocity v₃, then indicate the correct v₃:v₁ratio.

a. 3:1

b. 2:1

c. 1:3

d. 1:2

Ans: c. 1:3

Solution:In the Bohr model, the velocity of an electron in an orbit is inversely proportional to the radius of that orbit for a given atom or ion. Specifically, the velocity v of an electron in the nth orbit of a hydrogen-like atom (or ion) is given by:

v_n= k/n

Where k is a constant for a particular atom/ion and n is the principal quantum number of the orbit (1 for the first orbit, 2 for the second, etc.).

From this relationship, it follows that the velocity of the electron decreases as the principal quantum number increases. Specifically, for the electron in the first Bohr orbit (n = 1) and the third Bohr orbit (n = 3) of the He⁺ion, the velocities v₁and v₃would be directly related by the ratio of their orbit numbers.

Thus: v₃/v₁= (k/3)/(k/1) = 1/3

Therefore, the velocity ratio v₃:v₁is 1:3 This indicates that the electron in the third orbit moves at one-third the velocity of the electron in the first orbit.

Q5. Which hydrogen-like species will have the same radius as that of the 1^stBohr orbit of the hydrogen atom?

a. n = 2, Li²⁺

b. n = 2, Be³⁺

c. n= 2, He⁺

d. n =3, Li²⁺

Ans: b. n = 2, Be³⁺

Q6. The number of unpaired electrons in Mn²⁺ion is:

a. 2

b. 3

c. 5

d. 6

Ans: c. 5

Solution:

The atomic number of Mn is 25.

Electronic configuration of Mn = [Ar]3d⁵4s²

Electronic configuration of Mn²⁺= [Ar]3d⁵

There are five 3d orbitals, each orbital occupied by a single electron. Thus, number of unpaired electrons is 5.

Mn²⁺= [Ar]3d⁵

Q7. The equivalent weight of Na2S2O3 (Gram molecular weight = M) in the given reaction I₂+ 2Na₂S₂O₃= 2NaI + Na₂S₄O₆is:

a. M/2

b. M

c. 2M

d. M/4

Ans: b. M

Solution:

The equivalent weight of a substance is the weight of the substance that reacts with or is equivalent to one mole of hydrogen ions or one mole of electrons. In this reaction, the oxidation state of sulfur in Na₂S₂O₃is +2, while in Na₂S₄O₆it is +2.5 This means that each sulfur atom in Na₂S₂O₃loses 0.5 electrons. Since there are two sulfur atoms per molecule of Na₂S₂O₃, the total number of electrons lost per molecule is 1. Therefore, the equivalent weight of Na₂S₂O₃is equal to its molecular weight divided by the number of electrons lost per molecule, which is M/1 or simply M.

Q8. The compressibility factor for a van der Waal gas at high pressure is:

a. 1 + RT/Pb

b. 1+ Pb/RT

c. 1 - Pb/RT

d. 1

Ans: b. 1+ Pb/RT

Q9. For a first-order reaction with a rate constant k, the slope of the plot of log (reactant concentration) against time is:

a. k/2.303

b. k

c. -k/2.303

d. -k

Ans: c. -k/2.303

Q10. A solution containing 4g of polymer in a 4.0-litre solution at 27°Cshows an osmotic pressure of 3.0 x 10^-4atm. The molar mass of the polymer in g/mol is:

a. 820000

b. 82000

c. 8200

d. 820

Ans: b. 82000

Quick Links:To further improve your overall preparation and chances of scoring 45+ in the Chemistry paper (Paper II), you should solveWBJEE 2025 mock testfrom the Physical Chemistry chapter. This will help you assess your progress and exam readiness, as well as boost your confidence and time management skills. A combination of WBJEE 2025 Physical Chemistry practice questions with solutions and mock test series is the key to success.

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