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BITSAT Chemistry Test - 5 - JEE MCQ


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30 Questions MCQ Test - BITSAT Chemistry Test - 5

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BITSAT Chemistry Test - 5 - Question 1

The correct order of the bond order of a and b is?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 1

The bond length is inversely proportional to the bond order. 
In the given compound:

Due to resonance, there is a delocalisation of pi-electrons so 'a' acquires a double bond character. So, the bond length of 'a' decreases while the bond order increases, i.e., the bond order of double bond is 2
2. Due to the partial double bond, 'a' has a bond order which is greater than one.
'b' is a single bond between oxygen and carbon. Bond order of a single bond is 1.

BITSAT Chemistry Test - 5 - Question 2

The C−C bond length of the following molecules is in the order:

Detailed Solution for BITSAT Chemistry Test - 5 - Question 2

Bond length is inversely proportional to the bond order. When more electrons participate in a bond formation, the bond becomes shorter.
By approximation, the bond lengths of two different atoms are the sum of the two individual covalent radii. ​​​
Greater the s− s - character in the orbital, stronger and shorter will be the bond.
The bond length of carbon-carbon triple bond is shorter, due to more % s− character (50%)  than carbon-carbon double bond with % s− character as 33% .
Bond length of −C=C− is shorter than resonance bond in benzene, which has a bond length, between single bond and double bond and is shorter than carbon-carbon single bond, which has 25% s−  - character. Therefore,C2H6>C6H6>C2H4>C2H2

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BITSAT Chemistry Test - 5 - Question 3

Which of the following statements is correct for electromeric effect?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 3

Electromeric effect is the complete shifting of π electrons present between two atoms in the presence of an attacking reagent. This effect is shown by compounds which have multiple bonds like:C=C, C≡C, N=N, C=O, C≡N etc.
It means π-bond is involved in this effect.
Since this effect takes place in the presence of an attacking reagent, it is a temporary effect. It is of two types i.e., + E and - E.
Positive Electromeric effect

Negative Electromeric effect

BITSAT Chemistry Test - 5 - Question 4

Which of the following does not show geometrical isomerism?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 4

Geometrical isomerism is shown by compounds having at least one C = C and the two groups attached to each carbon must be different.

So, 1,1-dichloro-1-pentene does not show geometrical isomerism due to the presence of two similar
Cl atoms on the same C-atom.

BITSAT Chemistry Test - 5 - Question 5

Mark the incorrect statement in nitrogen Kjeldahl's method of estimation

Detailed Solution for BITSAT Chemistry Test - 5 - Question 5

In case of kjeldahl's method the percentage of  N2 is then calculated from the amount of NH3

BITSAT Chemistry Test - 5 - Question 6

Which behaves both as a nucleophile as well as an electrophile?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 6

CH3NH2 and CH3OH are nucleophiles, CH3−Cl is an electrophile. But  is a nucleophile due to the presence of a lone pair of electrons on N and is an electrophile due to the presence of a partial positive charge on C.

BITSAT Chemistry Test - 5 - Question 7

Mesomeric effect does not operate in:

Detailed Solution for BITSAT Chemistry Test - 5 - Question 7

Mesomeric effect: It is the movement of the pi -electrons in the conjugated system.
Conjugated system (for carbocations):- Means alternate vacant p- orbital and double bonds. 
The carbocation present adjacent to the oxygen atom is stabilised by the +M effect of the oxygen or any other hetero atom (which has a lone pair), and involves donation of lone pair of electrons to the vacant p− orbital of the carbon.
In the option (2) , the vacant p- orbital of the carbon atom and the lone pair of oxygen are not adjacent to each other. 
In option (1) the vacant p- orbital of the carbon atom and the lone pair of oxygen are adjacent to each other, so, the positive charge is stabilised by + M effect. 
In option (3), the​​​​​ vacant p- orbital of the carbon atom and the lone pair of nitrogen are adjacent to each other, so, the positive charge is stabilised by + M effect.
In option (4), the positive charge is stabilised by +M effect of the three phenyl groups. 

BITSAT Chemistry Test - 5 - Question 8

The correct order of stability for the following alkoxides is:

Detailed Solution for BITSAT Chemistry Test - 5 - Question 8

Here, in this case, we have to determine the stability of alkoxides. 
Here C  is more stable because in this case -ve charge on O is in conjugation with a double bond and NO2 group​. So there is an extra double bond in conjugation.
In case of B, -ve charge on oxygen is in conjugation with a double bond and not with NO2 group.
In case of A, there is no double bond so, no conjugation possible. So, least stable.
So, stability order is C>B>A


BITSAT Chemistry Test - 5 - Question 9

Arrange the following carbocations according to their stability order

Detailed Solution for BITSAT Chemistry Test - 5 - Question 9

Resonance and inductive effect decide stability of carbocations.

∴  Correct order of stability is
(II) <(I) <(III) <(IV)

BITSAT Chemistry Test - 5 - Question 10

2 gm of a mixture of CO and CO2 on reaction with excess of I2O5 yields 2.54 gm of I2 . What would be the mass percentage of CO2 in the original mixture?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 10

Let in the given mix. Mass of CO = x gm

I2O5+5CO ⟶ 5CO2+I2
1 mole Iis produced from 5 moles of CO
=0.01 mole Iis produced from 0.05 moles of CO.

BITSAT Chemistry Test - 5 - Question 11

What is the maximum amount of BaSO4 that can be precipitated by mixing 0.5 M M BaCl2 with 1 M H2SO4?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 11



BaCl2 is the limiting reagent. So, the amount of BaSO4 will correspond to the amount of BaCl2.

BITSAT Chemistry Test - 5 - Question 12

Two oxides of a metal contain 50% and 40% metal (M) respectively. If the formula of first oxide is MO2, the formula of second oxide will be

Detailed Solution for BITSAT Chemistry Test - 5 - Question 12


As first oxide is MO2
Let atomic mass of M = x

Or 0.5 x x + 16 = 32
Or 0.5 x = 16
x=32
∴ At. Mass of metal  m = 32
Let formula of second oxide is M2On

Therefore, formula of second oxide = M2O6
 Or =MO3

BITSAT Chemistry Test - 5 - Question 13

2g of O2 at 0C and 760 mm of Hg pressure has volume

Detailed Solution for BITSAT Chemistry Test - 5 - Question 13

Given mass of O2=2 g at 0℃ and 760 mm Hg
32 g of O2 = 22.4 L at STP
∴ 2g of O2 = x 2 = 1.4L

BITSAT Chemistry Test - 5 - Question 14

The atomic weights of two elements A and B are 40 u and 80 u, respectively. If xg of A contains y atoms, how many atoms are present in 2x g of B?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 14

One mole of any element contains NA entities.
Number of atoms in
Number of atoms in

BITSAT Chemistry Test - 5 - Question 15

Carbon dioxide contains 27.27 % of carbon, carbon disulphide contains 15.79% of carbon and sulphur dioxide contains 50% of sulphur. This data is an agreement with:

Detailed Solution for BITSAT Chemistry Test - 5 - Question 15

In CS2
C : S mass ratio is 15.79 : 84.21
15.79 parts of carbon combine with sulphur = 84.21
∴ 27.27 parts of carbon will combine with


Hence, ratio of S : O is 145.434:72.73 ie, 2 : 1
In SO2, the ratio of S : O is 1 : 1
Since, the ratio of S : O is a simple whole number ratio,
Therefore, law of reciprocal proportions is proved.

BITSAT Chemistry Test - 5 - Question 16

The reactant which is entirely consumed in the reaction is known as limiting reagent. In the reaction,
 2A+4B⟶3C+4D,
when 5 moles of A react with 6 moles of B then, calculate the amount of C formed.

Detailed Solution for BITSAT Chemistry Test - 5 - Question 16

Divide the no. of moles of each reactant by stoichiometric co - efficient in the balanced equation. Which ever provides least quotient is limiting reagent.

i.e., B is limiting reagent & product side calculations based on this
 4 moles of B → 3 moles of C
6 moles B → 3/4 x 6 = 4.5 moles of C.

BITSAT Chemistry Test - 5 - Question 17

The maximum number of moles of Ba3 (PO4)2 that can be formed if 2 mole BaCl2 is mixed with 1 mole Na3 PO4 is

Detailed Solution for BITSAT Chemistry Test - 5 - Question 17

The balanced chemical reaction is
3BaCl2+2Na3PO4→Ba3(PO4)2+6NaCl
According to the reaction,
3 mole BaCl2 reacts with 2 mole Na3 (PO4)2
Thus, 2 mole BaCl2 will requiremole Na3 PO4
Since sodium phosphate is taken in less quantity then the required quantity, here, Na3PO4 is the limiting reactant. The stoichiometric calculations are done based on the limiting reagent.
According to the reaction.
2 mole Na3 PO4 gives 1 mole Ba3(PO4)2
So,1 mole Na3PO4 will give 0.5 mole Ba3(PO4)2.

BITSAT Chemistry Test - 5 - Question 18

A compound is composed of 74% C , 8.7% H and 17.3% N by mass. If the molecular mass of the compound is 162, what is its molecular formula?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 18


Hence, the simplest molecular formula(empirical formula) C5H7N
The empirical Mass = 81
Now, molecular formula can be calculated using the ratio of molecular mass and empirical mass.
The ratio of molecular mass and empirical mass = 2 : 1
Hence, the molecular formula of the compound is C10H14N2.

BITSAT Chemistry Test - 5 - Question 19

In three moles of ethane (C2H6), calculate the following

(i) Number of moles of carbon atoms.
(ii) Number of moles hydrogen atoms.
(iii) Number of molecule of ethane.

Detailed Solution for BITSAT Chemistry Test - 5 - Question 19

(i) 1 mole of ethane, C2H6 contains 2 moles of carbon atoms.
∴ 3 moles of ethane, C2H6 will contain 2 × 3 = 6 moles of C-atoms.
(ii) 1 mole ethane, C2H6 will contains 6 moles of H-atoms.
∴ 3 moles of ethane, C2H6 will contain 3 × 6 = 18 moles of H-atoms.
(iii) 1 mole of ethane, C2H6 = 6.022 × 1023 molecules of ethane.
∴ 3 moles of ethane,C2H6= 3 × 6.022 × 1023 = 18.066 × 1023 molecules of ethane.

BITSAT Chemistry Test - 5 - Question 20

Which of the following is known as alpha particle ?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 20

Correct option is D.
An alpha particle is obtained by removing 2 electrons from a helium atom. So, an alpha particle is doubly - charged helium ion.

BITSAT Chemistry Test - 5 - Question 21

The number of atoms in a simple cubic unit cell are

Detailed Solution for BITSAT Chemistry Test - 5 - Question 21

The number of atoms in a simple cubic unit cell is one because contribution of atom at eight corners of the cube = 

BITSAT Chemistry Test - 5 - Question 22

In nuclear reactor, the control rods are made up of cadmium because cadmium

Detailed Solution for BITSAT Chemistry Test - 5 - Question 22

In nuclear reactors, the control rods are made up of cadmium because cadmium rod absorbs excess neutrons.

BITSAT Chemistry Test - 5 - Question 23

Which of the following is the composition of solder?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 23

Solder is an alloy of lead. It is made up of 33% of lead and 67% of Tin.

BITSAT Chemistry Test - 5 - Question 24

How many electrons can be accommodated in a p- orbital?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 24

P orbitals take a maximum of 6 electrons. 2px2, 2py2, 2pz2. The p orbitals are along 3 perpendicular axis. So p orbitals can accommodate a maximum of 6 electrons.

BITSAT Chemistry Test - 5 - Question 25

Which of the following shows iso-structural species?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 25

In SO42-, PO43- and BF4- all the central atoms are in sp3 - hybridisation state. Hence they are isostructural species (tetrahedral).

BITSAT Chemistry Test - 5 - Question 26

In aqua-regia the ratio of cone. HNO3 and cone. HCI present is

Detailed Solution for BITSAT Chemistry Test - 5 - Question 26

The aqua-regia is a mixture of one part of cone. HN03 and three parts of cone. HCI.

BITSAT Chemistry Test - 5 - Question 27

Absolute zero temperature is the temperature at which

Detailed Solution for BITSAT Chemistry Test - 5 - Question 27

Absolute zero is the temperature of which the velocities of the gas molecules reduce to zero. It means molecular motion ceases at absolute zero.

BITSAT Chemistry Test - 5 - Question 28

Which of the following product is formed by the reaction of sulphurdioxide with chlorine in presence of sunlight ?

Detailed Solution for BITSAT Chemistry Test - 5 - Question 28

In sunlight, sulphurdioxide and chlorine reacts as follows :

BITSAT Chemistry Test - 5 - Question 29

Surface water contains

Detailed Solution for BITSAT Chemistry Test - 5 - Question 29

The surface water contains suspended impurities.

BITSAT Chemistry Test - 5 - Question 30

 Avogadro’s number is the number of molecules present in

Detailed Solution for BITSAT Chemistry Test - 5 - Question 30

Avogadro’s number is 6.023 x 1023 which is equal to molecules present in gram molecular mass.

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