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Bihar PGT/TGT/PRT Exam  >  Bihar PGT/TGT/PRT Tests  >  Bihar STET Mock Test Series 2025  >  Bihar STET Paper 2 Chemistry Mock Test - 2 - Bihar PGT/TGT/PRT MCQ

Bihar STET Paper 2 Chemistry Mock Test - 2 - Bihar PGT/TGT/PRT MCQ


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30 Questions MCQ Test Bihar STET Mock Test Series 2025 - Bihar STET Paper 2 Chemistry Mock Test - 2

Bihar STET Paper 2 Chemistry Mock Test - 2 for Bihar PGT/TGT/PRT 2025 is part of Bihar STET Mock Test Series 2025 preparation. The Bihar STET Paper 2 Chemistry Mock Test - 2 questions and answers have been prepared according to the Bihar PGT/TGT/PRT exam syllabus.The Bihar STET Paper 2 Chemistry Mock Test - 2 MCQs are made for Bihar PGT/TGT/PRT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Bihar STET Paper 2 Chemistry Mock Test - 2 below.
Solutions of Bihar STET Paper 2 Chemistry Mock Test - 2 questions in English are available as part of our Bihar STET Mock Test Series 2025 for Bihar PGT/TGT/PRT & Bihar STET Paper 2 Chemistry Mock Test - 2 solutions in Hindi for Bihar STET Mock Test Series 2025 course. Download more important topics, notes, lectures and mock test series for Bihar PGT/TGT/PRT Exam by signing up for free. Attempt Bihar STET Paper 2 Chemistry Mock Test - 2 | 150 questions in 150 minutes | Mock test for Bihar PGT/TGT/PRT preparation | Free important questions MCQ to study Bihar STET Mock Test Series 2025 for Bihar PGT/TGT/PRT Exam | Download free PDF with solutions
Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 1
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What is cell entropy change of the following cell?

Pt(s) | H2(g) | CH3COOH, HCl || KCl (aq) |Hg2Cl2| (s) | Hg

P = 1 atm        0.1M                0.1M

Emf of the cell is found to be 0.045 V at 298 K and temperature coefficient is 

3.4 x10–4  VK–1

Given Ka (CH3COOH) = 10–5 M

Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 1

 

= 65.223J/K/mole

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 2
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Powdered substances are more effective adsorbents than their crystalline forms because:

Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 2

Greater the surface area, greater will be the area available for adsorption.

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Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 3
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Correct decreasing order of dipole moment of CH3F, CH3CI and CH3Br is

Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 3




Bond length and (EN) oppose each other, with the larger halogens having longer bonds but weaker (EN). Thus CH3CI > CH3F > CH3Br.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 4
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We can obtain ethylamine by Hoffmann bromamide reaction. The amide used in this reaction is:
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 4

The correct answer is option C
CH3​CH2​CONH2​ (A)⟶ CH3​ − CH2 ​− NH2​ (B)⟶ ​CH3​ − CH2 ​− OH
In the above sequence A & B respectively are Br2​/KOH and HNO2​
The first step is Hoffmann bromamide degradation reaction in which an amide (propanamide) is converted to an amine  (ethylamine) containing one carbon atom less. The reagent A is bromine in presence of KOH. In the second step, aliphatic primary amine (ethyl amine) reacts with nitrous acid (reagent B) to form aliphatic primary alcohol (ethyl alcohol).
 

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 5
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The rate constant of the reaction at temperature 200 K is 10 times less than the rate constant at 400 K. The activation energy of the reaction is  
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 5

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 6
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Consider the following transformations.

Q. Which reaction sequence will bent bring about the above transformation?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 6

On addition of H2/Pd/CaCO3
we get H-CH=C-CH2-CH2-CH3 
H-CH=C-CH2-CH2-CH3 + NBS/CCl4   →   H-CH=C-CHCl-CH2-CH3  (NBS adds Br to carbon at next alternate to double bond)
H-CH=C-CHCl-CH2-CH3 + KOH/EtOH  →  H2C=C-CH=CH-CH3
(by E2 elimination)

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 7
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The equilibrium constant for the reaction

CO(g) + H2O(g)  CO2(g) + H2(g) is 3 at 500 K. In a 2 litre vessel 60 gm of water gas [equimolar mixture of CO(g) and H2(g)] and 90 gm of steam is initially taken.

What is the equilibrium concentration of H2(g) at equilibrium (mole/L) ?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 7







Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 8
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What is known as Auto-oxidation?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 8
  • Autoxidation is any oxidation that occurs in presence of oxygen.
  • The term is usually used to describe the degradation of organic compounds in air (as a source of oxygen).
  • Autoxidation produces hydroperoxides and cyclic organic peroxides.
Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 9
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Two liquids X and Y form an ideal solution At 300 K, vapour pressure of the solution containing 1 mol of X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution, vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their pure states will be, respectively -  

[AIEEE 2009]
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 9

The correct answer is option B
Let 
P = xx​Px0 ​+ xy​Py0

on solving for Py0​ and Px0​
We get,
Px0​ = 400 mm of Hg
Py0 ​= 600 mm of Hg

*Answer can only contain numeric values
Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 10
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At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?

Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 10






Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 11
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Which among the following is adsorbed greatly by activated charcoal?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 11

SO2 due to high polarity and surface area is adsorbed maximum.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 12
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In oxygen difluoride (OF2) and dioxygen difluoride(O2F2), the oxygen is assigned an oxidation number of
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 12

Oxidation Numbers in Oxygen Difluoride (OF2) and Dioxygen Difluoride (O2F2)

Oxidation numbers are assigned to elements in compounds according to several rules. Here, we'll use the rule that states the oxidation number of fluorine (F) in a compound is always -1, and the rule that states the sum of oxidation numbers of all atoms in a compound is 0.

Oxidation Number in OF2:

  • In Oxygen Difluoride (OF2), there are two fluorine atoms, each with an oxidation number of -1. So, the total oxidation number contributed by fluorine is -2.
  • To make the total oxidation number 0, the oxygen atom must have an oxidation number of +2.


Oxidation Number in O2F2:

  • In Dioxygen Difluoride (O2F2), there are two fluorine atoms, each with an oxidation number of -1. So, the total oxidation number contributed by fluorine is -2.
  • There are also two oxygen atoms. To make the total oxidation number 0, the combined oxidation number of the two oxygen atoms must be +2.
  • Therefore, the oxidation number of each oxygen atom in O2F2 is +1 (because +2 divided by 2 equals +1).


So, in OF2 and O2F2, the oxygen is assigned an oxidation number of +2 and +1, respectively. Therefore, the correct answer is B: +2 and +1.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 13
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Which of the following types of octahedral complexes will exhibit geometrical isomerism (where, M = metal, a, b = achiral ligands)?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 13

Octahedral complexes of formula [Ma4b2] in which the two ligands b may be oriented cis or trans to each other, e.g.

Geometrical isomers (cis and trans) of [Co(NH3)4CI2]+.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 14
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Entropy change when 2 moles of an ideal gas expands reversibly from an initial volume of 1 dm3 to a final volume of 10 dm3 at a constant temperature of 298 K is
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 14

∆S= 2.303nR × log(V2/V1)Here n=2, R=8.314, V2= 10, V1= 1

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 15
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Comprehension Type

This section contains a passage describing theory, experiments, data, etc. Two questions related to the paragraph have been given. Each question has only one correct answer out of the given 4 options (a), (b), (c) and (d)

                                            Passage I

A constant current of 30 A is passed through an aqueous solution of NaCl for a time of 1.00 h.

Thus NaOH formed due to electrolysis is
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 15

The correct answer is option B
Formula used : W=ZIt
where Z is the equivalent mass of the deposit divided by 1 Faraday charge i.e. 96500 C.

Now, W= [(40/{+1})/96500] x 30 x  3600s

≈ 44.8 g

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 16
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Once the equilibrium is reached under given condition:
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 16

In a chemical reaction, chemical equilibrium is the state in which the forward reaction rate and the reverse reaction rate are equal. The result of this equilibrium is that the concentrations of the reactants and the products do not change. However, just because concentrations aren’t changing does not mean that all chemical reaction has ceased. Just the opposite is true; chemical equilibrium is a dynamic state in which reactants are being converted into products at all times, but at the exact rate that products are being converted back into reactants. The result of such a situation is analogous to a bridge between two cities, where the rate of cars going over the bridge in each direction is exactly equal. The result is that the net number of cars on either side of the bridge does not change.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 17
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Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl- (aq) is
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 17

½ H2(g) + ½ Cl2(g) → HCl(g) ∆fH° = -22.10 kcal mol-1  -----(I)
HCl(g) + H2O(l) → H+(aq) + Cl-(aq)     ∆solutionH° = -17.9 kcal mol-1 -----(II)
Let us add (I) and (II), we get
½ H2(g) + ½ Cl2(g) → H+(aq) + Cl-(aq) ∆rH° = -40.0 kcal mol-1
∆rH° = ∑∆Hproduct  -  ∑∆Hreactant = ∆fHCl- (since enthalpy of formation of H, H2 and Cl2 are 0)
∆fHCl- = -40.0

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 18
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The ≈ pH of the neutralisation point of 0.1 N ammonium hydroxide with 0.1 N HCl is
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 18

Salt formed : NH4Cl = 0.1 N
Solution will be slightly acidic due to Hydrolysis

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 19
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In an experiment conducted on a hungry chimpanzee, some bananas were kept outside its cage, but beyond its reach. Some sticks were also kept in its cage. After several unsuccessful attempts to reach out to the bananas, the chimpanzee pondered over the problem. Then, he picked up a stick and pulled the bananas towards itself. In this case, learning took place by
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 19

Animals solve problems by understanding – like human beings, instead of going through either the trial and error process or stimulus response association. In this case, learning took place by insight. Learning by insight involves finding a solution to the problem by thinking over the problem and understanding the various aspects of the problematic situation.
Insight learning refers to the sudden realisation of the solution of any problem without repeated trials or continuous practices. To further elaborate on its definition, insight learning is the type of learning in which one draws on previous experience and also seems to involve a new way of perceiving logical and cause-and-effect relationship.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 20
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_________ is the study of cognitive development that is concerned with basic mechanics of learning.
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 20

The behavioristic approach focuses on the mechanics of learning.

  • Behaviorist approach is the study of cognitive development that is concerned with the basic mechanics of learning. Learning - changes in behavior because of experiences (excluding changes due to fatigue, injury, or drug effects), is at the center of this approach.
  • The central tenet of behaviorism is that an organism learns behavioral adaptation, whose learning is governed by the principles of association (forming connections between ideas and events).
  • Behaviorism stands apart from other approaches by its emphasis on the relationship between observable behavior (responses) and environmental events (stimuli).

Behaviorism or the behaviorist approach lays stress on the role of environmental stimuli in determining the way we act. Behaviorism is also referred to as behavioristic psychology.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 21
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How does NEP 2020 propose to address the flexibility in subject choices for students in higher education?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 21

NEP 2020 suggests a flexible higher education system with multiple entry and exit points, allowing students to choose their courses based on their interests and career goals.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 22
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What is the purpose of the National Testing Agency (NTA) in the context of NEP 2020?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 22

NTA is responsible for conducting standardized university entrance exams as per the recommendations of NEP 2020.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 23
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Which of the following is a socialization agency?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 23

Socialization is an important process for the functioning and continuation of society. It is a process of learning the rules, habits, and values of a group to which a person belongs whether it is family, friends, colleagues, or any other group.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 24
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Madam Soni found that most of the students this year performed better in exams than the last year. This may be due to:
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 24

Madam Soni found that most of the students this year performed better in exams than the last year. This is because she improved the teaching methods, and thus, the students were able to understand better. This better understanding resulted in the improved performance of the students.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 25
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Assessment methods eventually improve
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 25

Assessment eventually improves student's learning. Student can improve his learning by knowing his mistakes through assessment. He can learn from his past mistakes and can improve his performance in the next assessment.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 26
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Mrs. Sharma can impart learning with better skill and results because she possesses
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 26

Mrs. Sharma can impart learning with better skill and results because she possesses excellent communication skills. Communication skills involve speaking, writing, imagery, body language, and the organisation of ideas into understandable structures. So, learning can be imparted in a better way because of these communication skills. Ability to interact with students in a better way reflects interpersonal skills. Knowledge of pedagogy helps in teaching by understanding pupils better.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 27
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Who has been appointed as the chairperson and independent director of PayU Payments Private Ltd?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 27

Renu Sud Karnad has been appointed as the chairperson and independent director of PayU Payments Private Ltd. Her extensive experience, including her tenure at HDFC Bank, highlights PayU's strategy of leveraging seasoned leadership to navigate the evolving fintech landscape.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 28
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In 1992, Rio Declaration was conducted on the issue of environment and development. In this conference, an action plan was framed for sustainable development. What was the name of that action plan?
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 28

Agenda 21 is an extensive arrangement of action to be made universally, broadly and locally by associations of the United Nations System, governments and major groups in each territory in which human affects the Earth.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 29
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Ramesh is studying about the various geographical barriers of the world. He also reads about the Palk Strait. The Palk Strait lies between
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 29

The Palk Strait is a strait between Tamil Nadu (India) and the Mannar area of the northern province of Sri Lanka.

Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 30
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The three equal circles touches each other externally. If the centres of these circles be A, B, C then ΔABC is ________
Detailed Solution for Bihar STET Paper 2 Chemistry Mock Test - 2 - Question 30

The figure is as follows:

Assuming, the radius of circle be r.
Then,
AB = BC = CA = 2r
So, all the three sides are equal of the triangle.
△ABC is an equilateral triangle.
Hence, the required answer is equilateral triangleequilateral triangle.

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