Mock Test - 1 - Class 12 MCQ

# Mock Test - 1 - Class 12 MCQ

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## 30 Questions MCQ Test - Mock Test - 1

Mock Test - 1 for Class 12 2024 is part of Class 12 preparation. The Mock Test - 1 questions and answers have been prepared according to the Class 12 exam syllabus.The Mock Test - 1 MCQs are made for Class 12 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Mock Test - 1 below.
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Mock Test - 1 - Question 1

### How many atoms of He would occupy the same volume as molecules of N2O4 , under similar temperature and pressure conditions?

Detailed Solution for Mock Test - 1 - Question 1

He is a monoatomic gas, So, it will have equal no. of atoms as that of N2O4 molecules

Mock Test - 1 - Question 2

### Three gases X, Y and Z has the van der Waals’ gas constant values ‘a’ as 1.42, 3.85 and 2.26 litre2 atm mol-2. The tendency of getting liquified is

Detailed Solution for Mock Test - 1 - Question 2

Vander Waal's constant ‘a’ represents, intermolecular attraction. So gas with higher value of ‘a’ can be easily liquefied.
∴ The order of gases will be Y > Z > X

Mock Test - 1 - Question 3

### 1 mole of chlorine gas in hot and concentrated alkaline medium disproportionates to

Detailed Solution for Mock Test - 1 - Question 3

In the above balanced reaction, it can be seen that 3 moles of Cl2 gives 5 moles of Cl- and 1 mole of ClO3-
∴ 1 mole of Cl will give 5/3 moles of Cl- and 1/3 moles of ClO3-

Mock Test - 1 - Question 4

In which of the following the cations occupy alternate tetrahedral voids in a cubic close packed arrangement ?

Detailed Solution for Mock Test - 1 - Question 4

ZnS crystallizes as a cubic close-packed array of S2- ions with Zn+2 ions in tetrahedral holes. If S2- ions occupy the lattice points of a face-centered cubic unit cell and Zn+2 ions are packed into every other tetrahedral hole, we get the unit cell of ZnS.

Mock Test - 1 - Question 5

What will be the van’t Hoff factor of the electrolyte in a solution having completely ionised complex salt sodium tetracyanozincate (II) ?

Detailed Solution for Mock Test - 1 - Question 5

For strong electrolytes, van't Hoff factor 'i' equals to total number of ions in the formula unit.

∴ From the above reaction it can be seen that sodium tetra cyanozincate (II) liberates 3 ions on complete ionisation.

Mock Test - 1 - Question 6

The solubility product has been recorded as 4 * 10-8 for BaF2. What will the solubility of BaF2 in a 0.1M HF solution ?

Detailed Solution for Mock Test - 1 - Question 6

Mock Test - 1 - Question 7

Some electrode potential data is given below

Fe3+ + e- → Fe2+ ; E0 = 0.77V

Zn2+ + 2e- → Zn; E0 = -0.76V

Br2 + 2e- → 2Br-; E0 = 1.08V

Ag+ + e- → Ag; E0 = 0.80V

According to the data given above, the reducing ability of various species can be arranged as

Detailed Solution for Mock Test - 1 - Question 7

Electrode with least E0 has highest reducing power of chemical species on the left.
∴ Order of reducing ability is Zn > Fe2+ > Ag > Br-

Mock Test - 1 - Question 8

Suggest the set of reagents that can bring about the following conversion

Detailed Solution for Mock Test - 1 - Question 8

Final product of the reaction is methyl amine, intermediate product must be isocyanide as given in the following reaction.

Mock Test - 1 - Question 9

[A] and [B] are respectively

Detailed Solution for Mock Test - 1 - Question 9

Lindlar’s catalyst gives syn addition of H-atoms & Na in Liq. NH3 gives anti-addition.
So, A → Cis-but-2-ene & B → Trans-but-2-ene

Mock Test - 1 - Question 10

Elimination of bromine from 2-bromo butane results in the formation of:

Detailed Solution for Mock Test - 1 - Question 10

Elimination of 20 halides is governed by Saytzeff’s rule, so, 2-Butene dominates here.

Mock Test - 1 - Question 11

Which of the following is a false remark ?

Detailed Solution for Mock Test - 1 - Question 11

Glucose is a monosaccharide as it can not be further hydrolyzed.

Mock Test - 1 - Question 12

The main reason for lanthanoid contraction is:

Detailed Solution for Mock Test - 1 - Question 12

Lanthanide contraction is the gradual decrease in the atomic and ionic size of lanthanoids with an increase in atomic number.
Causes of lanthanide contraction

• With an increase in the atomic number, the positive charge on nucleus increases by one unit and one more electron enters same 4f subshell.
• The electrons in 4f subshell imperfectly shield each other. Shielding in a 4f subshell is lesser than in d subshell.
• With the increase in nuclear charge, the valence shell is pulled slightly towards the nucleus. This causes lanthanide contraction.
Mock Test - 1 - Question 13

Which one of the following reactions examplify the oxidising property of hydrogen peroxide ?

Detailed Solution for Mock Test - 1 - Question 13

When H2O2 acts as oxidising agent, then H2O→ H2O + [O]

In reaction with Cu2S, S is oxidised to SO2 by nascent oxygen.

Mock Test - 1 - Question 14

The multiplicity of electrons in 3d5 electronic configuration is equal to

Detailed Solution for Mock Test - 1 - Question 14

Spin multiplicity is 2S+1, where S is total spin

spin multiplicity =

Mock Test - 1 - Question 15

KF combines with HF to form KHF2. The compound contains the species

Detailed Solution for Mock Test - 1 - Question 15

Fion has ability to form H-bonds with HF ; KHF2 → K+ + HF2-

Mock Test - 1 - Question 16

If the intermolecular forces vanish away, the volume occupied by the molecules contained in 4.5 kg water at STP will be

Detailed Solution for Mock Test - 1 - Question 16

When intermolecular forces vanish, water liquid charge to gas, So, mass & moles remain same but volume changes

Mock Test - 1 - Question 17

If uncertainty in position of electron is zero, the uncertainty in its momentum would be

Detailed Solution for Mock Test - 1 - Question 17

According to Heisenberg's Uncertainty Principle, it is impossible to measure position x and momentum p simultaneously.
Using Δx.Δp ⩾ h/4π
as Δx→0, Δp→∞
Here, neither uncertainty can be zero and if the position is zero the uncertainty in momentum becomes infinite.

Mock Test - 1 - Question 18

Which of the following is disproportionation reaction ?

Detailed Solution for Mock Test - 1 - Question 18

Disproportionation involves oxidation and reduction of atoms of the same element from same molecular species. Given reaction is cannizarro’s reaction.

Mock Test - 1 - Question 19

Number of H+ ions present in 10 mL of a solution of pH = 3 are

Detailed Solution for Mock Test - 1 - Question 19

Given pH = 3, Volume = 10 ml
∴ [H+] = 10-3 M
No. of moles of H+ in 10 ml is ′x′.
= 10-3 × 10-3 × 10
=10-5
No. of [H+] gms in 1 mole = 6.023×1023
Hence, no. of H+ ion in 10-5 moles = 6.023×1023×10-5
= 6.023×1018

Mock Test - 1 - Question 20

During developing of an exposed camera film, one step involves the following reaction,

Which of the following best describes the role of hydroquinol, it acts as

Detailed Solution for Mock Test - 1 - Question 20

It acts as reducing agent as, hydroxy group is oxidised & silver is reduced

Mock Test - 1 - Question 21

The data given below are for vapour phase reactions at constant pressure

The enthalpy change for the reaction

Detailed Solution for Mock Test - 1 - Question 21

Subtracting reaction (1) from reaction (2), We get

∴ Enthalpy change of the reaction is -252 KJ Mol-1

Mock Test - 1 - Question 22

Which of the following oxides is most acidic ?

Detailed Solution for Mock Test - 1 - Question 22
• Acidic nature of oxides increases with increases in non-metallic character or oxidation state.
• More the positive oxidation state of a non-metallic oxide more is its acidic nature.
Mock Test - 1 - Question 23

Treatment of propionaldehyde with dil. NaOH solution gives

Detailed Solution for Mock Test - 1 - Question 23
• Propionaldehyde will undergo an aldol condensation reaction when it is treated with dilute NaOH (base).
• The reaction results in the formation of 3-hydroxy-2-methylpentanal.
• This reaction initially involves the formation of the enolate ion which attacks another mole of the aldehyde to form the beta-hydroxyaldehyde product.

The reaction is given as follows:

Mock Test - 1 - Question 24

Which of the following reactions will yield 2,2-dibromopropane ?

Detailed Solution for Mock Test - 1 - Question 24

Reaction of propyne with 2 molecules of HBr to form 2,2 -dibromo propane can be written as:
CH3​−C≡CH + 2HBr → CH3​−C(Br)=CH2
CH3​−C(Br)=CH2 + HBr → CH3−C(Br)2−CH3
The reaction follows Markownikoff's rule.

Mock Test - 1 - Question 25

In [Ni(CO)4] complex Ni-C bond have

Detailed Solution for Mock Test - 1 - Question 25

[Ni(CO)4 has synergic bonding, CO is a π- acid ligand. Compound has C → Metal, sigma bond & Metal → C, π- bond.

Mock Test - 1 - Question 26

Chromyl chloride test is not given by

Detailed Solution for Mock Test - 1 - Question 26
• Chromyl chloride test is done for detecting the presence of Cl- ions.
• The chlorides of silver, lead, mercury and antimony are covalent in nature and thus do not generate Cl- ions and so they do not give the chromyl chloride test also.
• So, Heavy metal chlorides don't give this test because they are not ionic.
Mock Test - 1 - Question 27

For the following electrochemical cell

EMF of the cell will increase

Detailed Solution for Mock Test - 1 - Question 27

dmg forms stable co-ordination compound with Ni2+. So, concentration of metal ion at anode decreases and emf increases.

Mock Test - 1 - Question 28

Consider the following statements

(i) First order reaction completes in infinite time. This true and a basic of reaction kinetics
(ii) Average life is independent of concentration for first order reactions t = 1/λ
(iii) t75% = 1.5 t1/2 for zero order reaction as there is linear dependence
(iv) t75% = 2t1/2 for first order reaction as there is exponential dependence

Correct among the following are

Detailed Solution for Mock Test - 1 - Question 28

To determine the correct statements among the given options, let's analyze each statement one by one:
(i) First order reaction completes in infinite time:
- This statement is true. A first-order reaction does not complete in a finite amount of time but continues indefinitely. The reaction rate decreases exponentially with time, but it never reaches zero.
(ii) Average life is independent of concentration for first order reactions t = 1/λ:
- This statement is true. The average life of a reactant molecule is the reciprocal of the rate constant (λ) for a first-order reaction. It is independent of the initial concentration of the reactant.
(iii) t75% = 1.5 t1/2 for zero order reaction as there is linear dependence:
- This statement is false. For a zero-order reaction, the time it takes for the concentration of the reactant to decrease by 75% (t75%) is equal to 3 times the half-life (t1/2). Therefore, t75% = 3t1/2.
(iv) t75% = 2t1/2 for first order reaction as there is exponential dependence:
- This statement is false. For a first-order reaction, the time it takes for the concentration of the reactant to decrease by 75% (t75%) is equal to approximately 1.386 times the half-life (t1/2). Therefore, t75% ≈ 1.386t1/2.
From the analysis above, we can conclude that the correct statements among the given options are (i) and (ii). Therefore, the correct answer is option C: (i), (ii).
Mock Test - 1 - Question 29

Which of the following will be oxidised by HIO4?

(i) R - CO - CO - R
(ii) R - CO - CHOH - R
(iii) R - CHOH - CH2 - CHOH - R
(iv) R - CHOH - CHOH - R

Detailed Solution for Mock Test - 1 - Question 29

HIO4 oxidises following combinations at adjacent positions.

Mock Test - 1 - Question 30

Which of the following aryl amine is most difficult to diazotize?

Detailed Solution for Mock Test - 1 - Question 30

Due to strong electron-withdrawing effect of the - NO2 group, the nucleophilicity of the - NH2 is reduced and hence diazotisation becomes difficult.

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