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BITSAT Chemistry Test - 1 - JEE MCQ


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30 Questions MCQ Test BITSAT Mock Tests Series & Past Year Papers 2025 - BITSAT Chemistry Test - 1

BITSAT Chemistry Test - 1 for JEE 2024 is part of BITSAT Mock Tests Series & Past Year Papers 2025 preparation. The BITSAT Chemistry Test - 1 questions and answers have been prepared according to the JEE exam syllabus.The BITSAT Chemistry Test - 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BITSAT Chemistry Test - 1 below.
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BITSAT Chemistry Test - 1 - Question 1

The alcohol manufactured from water gas is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 1

Water gas and H2 reacts to produce methanol.

BITSAT Chemistry Test - 1 - Question 2

Two gm of oxygen at NTP occupies the volume

Detailed Solution for BITSAT Chemistry Test - 1 - Question 2

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BITSAT Chemistry Test - 1 - Question 3

Number of unpaired electrons in 1s2, 2s2 2p3 is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 3

1 s2 2s2 2p3 can be shown as 

Therefore, there are 3 unpaired electrons.

BITSAT Chemistry Test - 1 - Question 4

When primary amine is heated with CS2 in presence of excess of mercuric chloride, it produce isothiocyanate. This reaction is known as

Detailed Solution for BITSAT Chemistry Test - 1 - Question 4

Primary amine on heating with CS2 is presence of excess of HgCI2 produce ethyl isothiocyanate which has smell of mustard oil. Therefore, it is known as Hoffmann’s mustard oil reaction.

BITSAT Chemistry Test - 1 - Question 5

The most stable carbonium ion is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 5

The most stable carbonium ion is triphenyl methyl carbonium ion because the π-electrons of three benzene rings are delocalized with tb vacant π-orbital of central carbon atom. Therefore, it is stabilized by resonance.

BITSAT Chemistry Test - 1 - Question 6

Total number of electrons in all the p-orbitals of bromine will be

Detailed Solution for BITSAT Chemistry Test - 1 - Question 6

The electronic configuration of bromine atom is
1s2, 2s2 2p6, 3s2 3p6 3d10,4s2 4p5
Hence, total number of p-electrons is 17.

BITSAT Chemistry Test - 1 - Question 7

Flux is used to remove

Detailed Solution for BITSAT Chemistry Test - 1 - Question 7

Flux is used to remove silica and undesirable metal oxide. It is an external material added during smelting to convert infusible impurities into an easily fusible material known as slag.

BITSAT Chemistry Test - 1 - Question 8

Ionization constant of acetic acid is 1.8 * 10-5. The concentration of H+ ions in 0.1 M solution is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 8

Since, acetic acid is a weak electrolyte, hence

BITSAT Chemistry Test - 1 - Question 9

Acetone and acetaldehyde can be identified by

Detailed Solution for BITSAT Chemistry Test - 1 - Question 9

Acetaldehyde and acetone both comes under carbonyl group. Both respond to 2, 4-dinitrophenyl hydrazine to produce yellow or orange precipitate. Aldehydes produce colour with Schiff’s reagent while ketones do not respond to this test.

BITSAT Chemistry Test - 1 - Question 10

On heating one end of a piece of metal, tlhe other end becomes hot because of

Detailed Solution for BITSAT Chemistry Test - 1 - Question 10

On heating the one end of the metal piece the other end becomes hot this is due to the movement of energised electrons from one end to other end.

BITSAT Chemistry Test - 1 - Question 11

From the rate expression for the following reaction, determine its order of reaction and the dimensions of the rate constant.
H2O2(aq)+3I(aq)+2H+⟶2H2O(l)+I3;Rate=k[H2O2][I]

Detailed Solution for BITSAT Chemistry Test - 1 - Question 11

Rate = k[H2O2] [l−]
Order of reaction =1 + 1 = 2
Dimensions of 'k'= 

BITSAT Chemistry Test - 1 - Question 12

An archaeologist unearths a bone sample and wants to know the age of the bone. Her chemist friend determines the 45.3% of the initial amount of carbon-14 is present in the bone sample. If the half-life of carbon-14 is 5730 yrs, calculate the age (in years) of the bone.

Detailed Solution for BITSAT Chemistry Test - 1 - Question 12

BITSAT Chemistry Test - 1 - Question 13

A substance 'A' decomposes by a first order reaction starting initially with [A]=2.00 M and after 200 min, [A] becomes 0.15 M . For this reaction, t1/2 is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 13

The given reaction started initially with [A] =2.00 M
Time taken, (t)=200 min and  [A] becomes 0.15 M
For a first order reaction, the rate constant is

(Where, a, a−x are the initial and final concentrations, respectively).

Further,

BITSAT Chemistry Test - 1 - Question 14

Arrhenius equation may not be represented as:

Detailed Solution for BITSAT Chemistry Test - 1 - Question 14

The Arrhenius equation gives the relations between rate constants at different temperatures as:

BITSAT Chemistry Test - 1 - Question 15

For the first-order decomposition reaction of N2O5 , it is observed that,
(i)  N2O5(g)→2NO2(g)+1/2O2(g);
(ii)  2N2O5(g)→4NO2(g)+O2(g);

Which of the following is true?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 15

For the 1st reaction,


For the 2nd reaction,


or , 
Since the rate of disappearance of N2O5 is same in both reactions,
therefore, equation (i) = equation (ii)

Hence,  We have, k=2k'.

BITSAT Chemistry Test - 1 - Question 16

What should be the age of the fossil for meaningful determination of its age?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 16

As the half life of C14 is 5760 years, so a 6 year old fossil’s age can’t be determined.
Further this technique cannot be used to date objects older than 30,000 
years.
After this length of time the radioactivity is too low to be measured.

BITSAT Chemistry Test - 1 - Question 17

Two substances A (t1/2=5 min) and B (t1/2=15 min) are taken in such a way that initially [A]=4[B] . The time after which both the concentration will be equal is : (Assume that reaction is first order)

Detailed Solution for BITSAT Chemistry Test - 1 - Question 17

Ct=C0e−Kt
According to the question,



Alternative solution :

When [A] = [B]

BITSAT Chemistry Test - 1 - Question 18

In the start of summer, a given sample of milk turns sour at room temperature (27oC) in  48 hour . In a refrigerator at 2oC , milk can be stored three times before it sours.
For souring of milk, At 2oC , the reaction is three times slower than at 27oC . The activation energy of the souring of milk is (in kJ mol−1) .

Detailed Solution for BITSAT Chemistry Test - 1 - Question 18

At 2oC (275 K) , the reaction is three times slower than at 27oC (300 K) . This implies that for souring of milk.

Applying Arrhenius equation 

Therefore, Ea=30.146 (kJ mol−1).

BITSAT Chemistry Test - 1 - Question 19

A + B → Product
If concentration of A is doubled, rate increases 4 times. If concentration of A and B are doubled, rate increases 8 times. The differential rate equation of the reaction will be

Detailed Solution for BITSAT Chemistry Test - 1 - Question 19

Let the order with respect to A and B is x and y respectively.
Hence,
Rate r=[A]x[B]y ...(i)
On doubling the concentration of A, rate increases 4 times,
4r=[2A]x[B]y ... (ii)
From Eqs. (i) and (ii)

∴  x = 2
∴ Order with respect to A is 2
If concentration of A and B both are doubled,
8r=[2A]x[2B]y  ... (iii)
From Eqs. (i) and (iii), we get

2y=2
∴ y = 1
Hence , differential rate equation is

[Where, CA and CB = concentrations of A and B]

BITSAT Chemistry Test - 1 - Question 20

Correct expression for mass number A is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 20

Mass no. = no. of protons + no. of neutrons.

BITSAT Chemistry Test - 1 - Question 21

According to quantum mechanics ψ2(r) the wave function squared gives

Detailed Solution for BITSAT Chemistry Test - 1 - Question 21

ψ has no physical significance while ψ2 represnts the probability density of finding an electron.

BITSAT Chemistry Test - 1 - Question 22

Spin quantum number with two spin states of the electron represented by two arrows, ↑ (spin up) and ↓ (spin down) was introduced to account for

Detailed Solution for BITSAT Chemistry Test - 1 - Question 22

Explanation


  • Anomalous Zeeman Effect: The spin quantum number was introduced to account for the anomalous Zeeman effect, which is the splitting of emission lines into more lines than predicted by the magnetic quantum number. This effect occurs when the electron spin interacts with an external magnetic field, causing the energy levels to split further based on the spin states of the electron.

  • Representation of Spin States: The two spin states of the electron, represented by the arrows ↑ (spin up) and ↓ (spin down), help differentiate between the two possible orientations of the electron's intrinsic angular momentum. This distinction is crucial in understanding how the electron interacts with external magnetic fields and contributes to the splitting of emission lines.

  • Quantum Mechanical Explanation: In quantum mechanics, the spin quantum number is a fundamental property of particles like electrons, and it plays a significant role in determining the behavior of these particles in different physical situations. The introduction of the spin quantum number allows for a more comprehensive description of electron states and their interactions with external fields.


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BITSAT Chemistry Test - 1 - Question 23

Give IUPAC names of the following compound:

Detailed Solution for BITSAT Chemistry Test - 1 - Question 23

This is 1-phenyl propan-1-one.

BITSAT Chemistry Test - 1 - Question 24

The reagents to bring about the change from but – 2 – ene to ethanal  

Detailed Solution for BITSAT Chemistry Test - 1 - Question 24

Reductive Ozonolysis of alkene forms aldehyde or ketone.

BITSAT Chemistry Test - 1 - Question 25

Which of the following is correct?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 25

C2H5OH get oxidizes to CH3CHO and –COCH3 group is important for iodoform.

BITSAT Chemistry Test - 1 - Question 26

A strong base can abstract an α – hydrogen from

Detailed Solution for BITSAT Chemistry Test - 1 - Question 26

Because conjugate base that form will be stable in case of Ketone.

BITSAT Chemistry Test - 1 - Question 27

Which of the following reagents can be used to convert a carboxylic acid directly into its corresponding acid chloride derivative?

Detailed Solution for BITSAT Chemistry Test - 1 - Question 27

BITSAT Chemistry Test - 1 - Question 28

Commercial concentrated nitric acid is 15.6 M. To prepare 10 L of 6.0 M nitric acid from it,

Detailed Solution for BITSAT Chemistry Test - 1 - Question 28

This is a case of dilution.

Thus, 3.846 L of 15.6 M solution is diluted to 10.0 L by addition of 6.154 L of H20.

BITSAT Chemistry Test - 1 - Question 29

Mole fraction of C3H5(OH)3 in a solution of 36 gm of water and 46 gm of glycerine is 

Detailed Solution for BITSAT Chemistry Test - 1 - Question 29

BITSAT Chemistry Test - 1 - Question 30

AgNO3 sample is 85% by mass. To prepare 125 mL of 0.05 M AgNO3 solution, AgNO3 sample required is

Detailed Solution for BITSAT Chemistry Test - 1 - Question 30

125 mL of 0.05 M AgN03 = 125 x 0.05 millimoles 

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