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Test: Chemistry Minor Mock Test- 2 (March 19) - NEET MCQ


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30 Questions MCQ Test - Test: Chemistry Minor Mock Test- 2 (March 19)

Test: Chemistry Minor Mock Test- 2 (March 19) for NEET 2024 is part of NEET preparation. The Test: Chemistry Minor Mock Test- 2 (March 19) questions and answers have been prepared according to the NEET exam syllabus.The Test: Chemistry Minor Mock Test- 2 (March 19) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Chemistry Minor Mock Test- 2 (March 19) below.
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Test: Chemistry Minor Mock Test- 2 (March 19) - Question 1

The specific conductances of four electrolytes in ohm−1cm−1 are given below. Which one offers higher resistance to passage of electric current?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 1

Specific conductance = Cell constant/Resistance
∴ Specific conductance ∝ 1/Resistance 
∴ The one with the lowest resistance (order 10-9) will have the highest specific conductance.
Hence, option A is correct.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 2

The elements of group 14 are slightly more electronegative than group 13 elements because of

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 2

As we move along the period electronegativity increases due to increase in nuclear charge.

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Test: Chemistry Minor Mock Test- 2 (March 19) - Question 3

How many Na+ ions are present in 100 mL of 0.25 M of NaCl solution?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 3


NaCl → Na+ Cl
No. of moles of Na+ ions = 0.025
No. of Na+ ions = 0.025 x 6.023 x 1023 = 1.505 × 1022

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 4

For nth order reaction,

Graphs between log (rate) and log[A0] are of the type ([A0] is the initial concentration)

Lines P, Q, R and S are for the order 

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 4


n = order of the reaction =slope of the line
Larger the value of the slope of the line, larger the order.
Thus, P = 3, Q = 2, R = 1, S = 0.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 5

In the reaction of metallic cobalt placed in nickel sulphate solution, therein is a competition for release of electrons At equilibrium, chemical tests reveal that both Ni+2 (aq) and Co+2 (aq) are present at moderate concentrations. The result is that:

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 5

The reaction of metallic cobalt in a nickel sulfate solution involves a competition for the release of electrons. This means that the cobalt metal can react with the nickel ions in the solution, or the nickel can deposit on the cobalt metal.

At equilibrium, the reaction has balanced out with no net change in the concentration of the reactants and products. The fact that both Ni+2 (aq) and Co+2 (aq) are present at moderate concentrations at equilibrium signifies that neither forward nor reverse reactions are greatly favoured.

  • A: This option is incorrect because both reactants and products are present in moderate concentrations, indicating that neither is greatly favoured.
  • B: This statement is not correct either. Even though Co (s) and Ni+2 (aq) are part of the reaction, the fact that Co+2 (aq) is also present at moderate concentrations shows that they are not the only favoured species.
  • C: This option is also incorrect. Even though Co+2 (aq) and Ni (s) are part of the reaction, the fact that Ni+2 (aq) is also present at moderate concentrations shows that they are not the only favoured species.
  • D: This is the correct answer. When a reaction is at equilibrium, it means that the rate of the forward reaction equals the rate of the reverse reaction. Therefore, neither the reactants nor the products are greatly favoured. In other words, the concentrations of the reactants and products remain constant over time
Test: Chemistry Minor Mock Test- 2 (March 19) - Question 6

Chlorine, bromine and iodine when combined with oxygen, have oxidation numbers

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 6
  • Oxidation Numbers of Halogens: The halogens (Fluorine, Chlorine, Bromine, Iodine, and Astatine) are a group in the periodic table. They are known for their high electronegativity and hence, when they combine with almost all other elements, they tend to have an oxidation number of -1 as they gain one electron to achieve a stable electronic configuration.
  • Exception in the case of Oxygen: However, oxygen is more electronegative than all halogens except fluorine. So, when halogens (chlorine, bromine, and iodine) combine with oxygen, they tend to lose electrons to oxygen and hence, they show positive oxidation states.
  • Conclusion: Therefore, when chlorine, bromine, and iodine are combined with oxygen, they have an oxidation number of +1 or any positive number (depending on the number of oxygen atoms they are combined with). Hence, the correct answer is C: +1 or any positive number.
Test: Chemistry Minor Mock Test- 2 (March 19) - Question 7

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 7


Test: Chemistry Minor Mock Test- 2 (March 19) - Question 8

Product (B) will be : 

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 8

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 9

 In a free radical reaction, free radicals are formed at

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 9


Here, we can see that in the initiation and termination step, radical is formed.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 10

The rate equation for the reaction,

2A + B → C

is found to be, rate = k[A] [B]

Q. The correct statement in relation to this reaction is that the

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 10

Rate = k (A)[B]

The given reaction is first order in A and first order is B.
Thus, total order = 2

(a) Unit of k = cone 1 - n time -1 = conc-1 time-1 Thus, (a) is false.
(b)  of second-order reaction, thus (b) is false. 

Thus, (c) is correct.

Thus, value of k is dependent on the concentration of A and B. Thus, (d) is false.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 11

 Which among the following is a very unstable and reactive species:

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 11

Carbocation is most reactive as it lacks electrons. So it need an electron to attain stability.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 12

The function of Fe (OH)3 in the contact process is

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 12

Fe (OH)3 is used to remove arsenic impurities.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 13

 Which of the following ions is most stable?   

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 13

The three CH3 radicals attached to the positively charged carbon are electron repelling group. Therefore, these CH3 radicals exhibit +I effect (Inductive effect) and repel electrons towards the positively charged Carbon. This decreases the positive charge and makes the compound more stable.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 14

Among the following which is the strongest oxidizing agent?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 14

F2 is very strong oxidizing agent.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 15

When a zinc rod is kept in a copper nitrate solution what happens?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 15

When zinc is placed in copper nitrate solution the intensity of the blue colour is produced and copper iron is deposited on zinc.

  • This is a Redox reaction between zinc and an aqueous solution of copper nitrate occurring in a beaker.
Test: Chemistry Minor Mock Test- 2 (March 19) - Question 16

The rate law for a reaction between the substances A and 8 is given by

rate = k[A]n [B]m

If concentration of A is doubled and that of B is halved, the new rate as compared to the earlier rate would be 

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 16



Test: Chemistry Minor Mock Test- 2 (March 19) - Question 17

(Major ) Product (A) will be :  

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 17


Test: Chemistry Minor Mock Test- 2 (March 19) - Question 18

The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is:

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 18

The correct order of equivalent conductance at infinite dilution is KCl > NaCl > LiCl.

Anion is same (chloride ion) for all the species. Larger is the size of the cation, greater is the equivalent conductance at infinite dilution and vice versa.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 19

Mono-chloro product (inculding steroisomers)are :

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 19





Test: Chemistry Minor Mock Test- 2 (March 19) - Question 20

What is the molarity of a solution containing 10 g of NaOH in 500 mL of solution?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 20

No. of moles of NaOH = 10/40 = 0.25 mol

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 21

Which of the following statements appears to the first order reaction?

2N2O5(CCl4) → 4NO2(CCl4) + O2(g)




Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 21

(a)

I, II, V - represent first order equation

III, VI - zeroth order equation

IV - second order equation

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 22

What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 22


Mass of C2H5OH = 85 g
Molar mass of C2H5OH = 46 g/mol
nC2H5OH = 85/46 = 1.85 mol
Mass of water = 100 - 85 = 15 g
nH2O = 15/18 = 0.833 mol

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 23

Calculate the percentage composition of a solution obtained by mixing 200 g of a 20% and 300 g of a 30% solution by weight.

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 23

Mass of solute is obtained by multiplying % weight with mass of solvent and obtaining an answer divided by 100.
Mass of solute in 200g of 30% solution can be obtained as follows:
30% by weight indicates 30g of solute is present in 100g of solvent so 60g of solute will be present in 200g of solution.
20% by weight indicates 20g of solute is present in 100g of solvent so 60g of solute will be present in 300g of solution
So on mixing these both solutions, mass of solute is 60g +60g =120g
Mass of solution on mixing is 200g +300g =500g
Percent of solute can be obtained as follows:

24% percentage of solute in the final solution.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 24

The decomposition of hydrogen peroxide to form water and oxygen is an example of

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 24

A disproportionation reaction is a particular type of redox reaction in which a single compound is transformed into two different compounds, one being reduced and the other being oxidized.

The decomposition of hydrogen peroxide to form water and oxygen is a classic example of a disproportionation reaction.

*Multiple options can be correct
Test: Chemistry Minor Mock Test- 2 (March 19) - Question 25

Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 25

Determining Anodes when connected to the Standard Hydrogen Electrode

The Standard Hydrogen Electrode (SHE) is commonly used as a reference electrode in electrochemistry and has an electrode potential of 0V. The electrode with a lower reduction potential than the SHE will act as the anode (site of oxidation), while the electrode with a higher reduction potential than the SHE will act as the cathode (site of reduction).
A: Fe/Fe2+ E0= − 0.44
The reduction potential of this electrode is lower than that of the SHE, so it will act as the anode when connected to the SHE.
B: Al/Al3+ E0= - 1.66
The reduction potential of this electrode is lower than that of the SHE, so it will also act as the anode when connected to the SHE.
C: F2(g)/2F(aq) E0= + 2.87
The reduction potential of this electrode is higher than that of the SHE, so it will act as the cathode, not the anode, when connected to the SHE.
D: Cu/Cu2+ E0= + 0.34
The reduction potential of this electrode is higher than that of the SHE, so it will also act as the cathode, not the anode, when connected to the SHE.

Therefore, electrodes A: Fe/Fe2+ (E0= − 0.44) and B: Al/Al3+ (E0= - 1.66) will act as anodes when connected to the Standard Hydrogen Electrode.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 26

A compound X, of boron reacts with NH3 on heating to give another compound. Y which is called inorganic benzene. The compound X can be prepared by treating BF3 with Lithium aluminium hydride. The compounds X and Y are represented by the formulas.

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 26

BF3 reacts with LiAlH4 to form B2H6 which reacts with NH3NH3 in 1:2 ratio forming B3N3H6( Borazine , inorganic benzene

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 27

In general, nonmetals as compared to metals have

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 27

Non metals have high value of ionization energy while metals have comparatively low value of ionization energy.

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 28

What is known as Auto-oxidation?

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 28
  • 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.
Test: Chemistry Minor Mock Test- 2 (March 19) - Question 29

The standard reduction potentials E°, for the half reactions are as:

The emf for the cell reaction,

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 29

Test: Chemistry Minor Mock Test- 2 (March 19) - Question 30

 Heterolytic cleavage is a way to cleave the:

Detailed Solution for Test: Chemistry Minor Mock Test- 2 (March 19) - Question 30

Heterolytic fission, also known as heterolysis, is a type of bond fission in which a covalent bond between two chemical species is broken in an unequal manner, resulting in the bond pair of electrons being retained by one of the chemical species (while the other species does not retain any of the electrons from the bond pair). When a neutrally charged molecule undergoes heterolytic fission, one of the products will have a positive charge whereas the other product will have a negative charge.

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