CBSE 12 Chemistry Question Paper-2019 by Pavan | Practice Test to Test Your Knowledge
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  3. CBSE 12 Chemistry Question Paper-2019
CBSE 12 Chemistry Question Paper-2019

CBSE 12 Chemistry Question Paper-2019

This mock test includes actual CBSE Class 12 Chemistry board exam questions from the year 2019, helping students understand exam trends and practice real paper format

2025-08-05
Pavan Pavan
CBSE Class 12 Chemistry 2019 Grade 12

Duration

30 min

Questions

30

Marking

Negative

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Questions Preview

Out of NaCl and AgCl, which one shows Frenkel defect and why?

A
NaCl shows Frenkel defect due to the presence of smaller chloride ions.
B
AgCl shows Frenkel defect due to the presence of smaller silver ions.
C
NaCl and AgCl both show Frenkel defect.
D
Neither NaCl nor AgCl shows Frenkel defect.

Arrange the following in increasing order of boiling points: (CH3)3N, C2H5OH, C2H5NH2

A
(CH3)3N < C2H5NH2 < C2H5OH
B
C2H5NH2 < (CH3)3N < C2H5OH
C
(CH3)3N < C2H5OH < C2H5NH2
D
C2H5OH < (CH3)3N < C2H5NH2

Why are medicines more effective in colloidal state? OR What is the difference between an emulsion and a gel?

A
Medicines are more effective in colloidal state because they have a larger surface area.
B
Emulsions are stable mixtures of oil and water, while gels are solid-like substances formed from a liquid.
C
Medicines are less effective in colloidal state as they cannot dissolve easily.
D
Emulsions are liquids, while gels are solids that do not interact with water.

Define ambident nucleophile with an example.

A
An ambident nucleophile has two nucleophilic sites. Example: NO2–
B
An ambident nucleophile can donate electrons through one site only. Example: CN–
C
An ambident nucleophile has multiple bonds. Example: H2O
D
An ambident nucleophile is not reactive. Example: Cl–

What is the basic structural difference between glucose and fructose? OR Write the products obtained after hydrolysis of lactose.

A
Glucose is an aldose, whereas fructose is a ketose.
B
Glucose and fructose are both aldoses.
C
The products of lactose hydrolysis are glucose and galactose.
D
Glucose and fructose are the same in structure.

Write balanced chemical equations for the following processes: (i) XeF2 undergoes hydrolysis. (ii) MnO2 is heated with conc. HCl.

A
(i) XeF2 + H2O → Xe + 2HF (ii) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
B
(i) XeF2 + H2O → XeF4 + 2HF (ii) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
C
(i) XeF2 + H2O → Xe + 2HF (ii) MnO2 + HCl → MnCl2 + H2O
D
(i) XeF2 + H2O → Xe + 2HF (ii) MnO2 + 2HCl → MnCl2 + H2O

State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations.

A
Raoult’s law states that the partial vapor pressure of each volatile component is directly proportional to its mole fraction. Characteristics: 1) The total vapor pressure is the sum of the partial vapor pressures. 2) The solution follows ideal behavior.
B
Raoult’s law states that the vapor pressure is independent of the mole fraction of the solute. Characteristics: 1) The vapor pressure decreases with an increase in solute. 2) It does not follow ideal behavior.
C
Raoult’s law states that the vapor pressure is constant for all components. Characteristics: 1) The solution behaves like a non-ideal solution. 2) The vapor pressure increases with an increase in solute.
D
Raoult’s law is applicable only to non-volatile solutes. Characteristics: 1) The total vapor pressure is dependent on the temperature. 2) The vapor pressure is unaffected by solute.

For a reaction, 2H2O2 → 2H2O + O2, the proposed mechanism is as given below: (1) H2O2 + I– → H2O + IO– (slow), (2) H2O2 + IO– → H2O + I– + O2 (fast). (i) Write the rate law for the reaction. (ii) Write the overall order of reaction. (iii) Which one is the rate-determining step?

A
(i) Rate law = k[H2O2][I–] (ii) Overall order = 2 (iii) The rate-determining step is the first step.
B
(i) Rate law = k[H2O2][IO–] (ii) Overall order = 1 (iii) The rate-determining step is the second step.
C
(i) Rate law = k[H2O2]² (ii) Overall order = 2 (iii) The rate-determining step is the second step.
D
(i) Rate law = k[H2O2][I–][IO–] (ii) Overall order = 3 (iii) The rate-determining step is the first step.

When MnO2 is fused with KOH in the presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidizes KI to compound (C), whereas an acidified solution of compound (B) oxidizes KI to compound (D). Identify (A), (B), (C), and (D).

A
(A) MnO2, (B) MnO4–, (C) I2, (D) I–
B
(A) MnO4–, (B) Mn2+, (C) I2, (D) I–
C
(A) Mn2+, (B) MnO4–, (C) I–, (D) I2
D
(A) MnO4–, (B) Mn2+, (C) I2, (D) I2

Write the IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex.

A
IUPAC name: Bis(ethylenediamine)chloroplatinum(II). Geometrical isomers: Cis and Trans.
B
IUPAC name: Chlorobis(ethylenediamine)platinum(II). Geometrical isomers: Cis only.
C
IUPAC name: Ethylenediamine platinum chloride. Geometrical isomers: Trans only.
D
IUPAC name: Platinum chloride ethylenediamine. Geometrical isomers: Cis and Trans.

Out of [CoF6]3– and [Co(en)3]3+, which one complex is (i) paramagnetic, (ii) more stable, (iii) inner orbital complex, and (iv) high spin complex? (Atomic no. of Co = 27)

A
[CoF6]3– is paramagnetic, more stable, inner orbital, and high spin.
B
[Co(en)3]3+ is paramagnetic, less stable, outer orbital, and low spin.
C
[CoF6]3– is paramagnetic, less stable, outer orbital, and low spin.
D
[Co(en)3]3+ is paramagnetic, more stable, inner orbital, and high spin.

Write structures of compounds A and B in each of the following reactions: (i) KMnO4 – KOH → A; H3O+ → B. (ii) CrO3 → A; H2N–NH–CONH2 → B.

A
(i) A is MnO2, B is Mn2+. (ii) A is CrO3, B is urea.
B
(i) A is MnO4–, B is Mn2+. (ii) A is Cr2O7, B is urea.
C
(i) A is MnO4–, B is MnO2. (ii) A is Cr2O3, B is urea.
D
(i) A is Mn2+, B is MnO2. (ii) A is CrO3, B is urea.

The decomposition of NH3 on platinum surface is a zero-order reaction. If rate constant (k) is 4 × 10–3 Ms–1, how long will it take to reduce the initial concentration of NH3 from 0.1 M to 0.064 M?

A
Time = (0.1 – 0.064) / (4 × 10–3) = 9 seconds.
B
Time = (0.1 – 0.064) / (4 × 10–3) = 3 seconds.
C
Time = (0.1 – 0.064) / (4 × 10–3) = 8 seconds.
D
Time = (0.1 – 0.064) / (4 × 10–3) = 5 seconds.

What is the role of activated charcoal in gas masks?

A
Activated charcoal adsorbs harmful gases, preventing them from entering the lungs.
B
Activated charcoal filters out solid particles from the air.
C
Activated charcoal increases the oxygen levels in the air.
D
Activated charcoal neutralizes toxic gases.

What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

A
The charge on ferric oxide colloidal particles is positive, and the sol is represented as Fe2O3 (colloidal).
B
The charge on ferric oxide colloidal particles is negative, and the sol is represented as Fe3O4 (colloidal).
C
The charge on ferric oxide colloidal particles is positive, and the sol is represented as Fe3O4 (colloidal).
D
The charge on ferric oxide colloidal particles is neutral, and the sol is represented as FeO (colloidal).

An element crystallizes in an fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm–3. Calculate the number of atoms in 108 g of the element.

A
The number of atoms in 108 g of the element is 3 × 10^23 atoms.
B
The number of atoms in 108 g of the element is 2 × 10^23 atoms.
C
The number of atoms in 108 g of the element is 4 × 10^23 atoms.
D
The number of atoms in 108 g of the element is 1 × 10^23 atoms.

A 4% solution(w/w) of sucrose (M = 342 g mol–1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water.

A
The freezing point of the 5% glucose solution is 272.15 K.
B
The freezing point of the 5% glucose solution is 273.15 K.
C
The freezing point of the 5% glucose solution is 270.15 K.
D
The freezing point of the 5% glucose solution is 269.15 K.

When MnO2 is fused with KOH in the presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidizes KI to compound (C), whereas an acidified solution of compound (B) oxidizes KI to compound (D). Identify (A), (B), (C), and (D).

A
(A) MnO2, (B) MnO4–, (C) I2, (D) I–
B
(A) MnO4–, (B) Mn2+, (C) I2, (D) I–
C
(A) Mn2+, (B) MnO4–, (C) I–, (D) I2
D
(A) MnO4–, (B) Mn2+, (C) I2, (D) I2

Write the IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex.

A
IUPAC name: Bis(ethylenediamine)chloroplatinum(II). Geometrical isomers: Cis and Trans.
B
IUPAC name: Chlorobis(ethylenediamine)platinum(II). Geometrical isomers: Cis only.
C
IUPAC name: Ethylenediamine platinum chloride. Geometrical isomers: Trans only.
D
IUPAC name: Platinum chloride ethylenediamine. Geometrical isomers: Cis and Trans.

Out of [CoF6]3– and [Co(en)3]3+, which one complex is (i) paramagnetic, (ii) more stable, (iii) inner orbital complex, and (iv) high spin complex? (Atomic no. of Co = 27)

A
[CoF6]3– is paramagnetic, more stable, inner orbital, and high spin.
B
[Co(en)3]3+ is paramagnetic, less stable, outer orbital, and low spin.
C
[CoF6]3– is paramagnetic, less stable, outer orbital, and low spin.
D
[Co(en)3]3+ is paramagnetic, more stable, inner orbital, and high spin.

Write structures of compounds A and B in each of the following reactions: (i) KMnO4 – KOH → A; H3O+ → B. (ii) CrO3 → A; H2N–NH–CONH2 → B.

A
(i) A is MnO2, B is Mn2+. (ii) A is CrO3, B is urea.
B
(i) A is MnO4–, B is Mn2+. (ii) A is Cr2O7, B is urea.
C
(i) A is MnO4–, B is MnO2. (ii) A is Cr2O3, B is urea.
D
(i) A is Mn2+, B is MnO2. (ii) A is CrO3, B is urea.

For a reaction, 2H2O2 → 2H2O + O2, the proposed mechanism is as given below: (1) H2O2 + I– → H2O + IO– (slow), (2) H2O2 + IO– → H2O + I– + O2 (fast). (i) Write the rate law for the reaction. (ii) Write the overall order of reaction. (iii) Which one is the rate-determining step?

A
(i) Rate law = k[H2O2][I–] (ii) Overall order = 2 (iii) The rate-determining step is the first step.
B
(i) Rate law = k[H2O2][IO–] (ii) Overall order = 1 (iii) The rate-determining step is the second step.
C
(i) Rate law = k[H2O2]² (ii) Overall order = 2 (iii) The rate-determining step is the second step.
D
(i) Rate law = k[H2O2][I–][IO–] (ii) Overall order = 3 (iii) The rate-determining step is the first step.

A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited at the cathode of cell Y.

A
Time = 180 seconds. Mass of Zn deposited = 3.2 g.
B
Time = 200 seconds. Mass of Zn deposited = 2.5 g.
C
Time = 150 seconds. Mass of Zn deposited = 3.0 g.
D
Time = 100 seconds. Mass of Zn deposited = 4.0 g.

What is the role of activated charcoal in gas masks?

A
Activated charcoal adsorbs harmful gases, preventing them from entering the lungs.
B
Activated charcoal filters out solid particles from the air.
C
Activated charcoal increases the oxygen levels in the air.
D
Activated charcoal neutralizes toxic gases.

What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

A
The charge on ferric oxide colloidal particles is positive, and the sol is represented as Fe2O3 (colloidal).
B
The charge on ferric oxide colloidal particles is negative, and the sol is represented as Fe3O4 (colloidal).
C
The charge on ferric oxide colloidal particles is positive, and the sol is represented as Fe3O4 (colloidal).
D
The charge on ferric oxide colloidal particles is neutral, and the sol is represented as FeO (colloidal).

Write structures of compounds A and B in each of the following reactions: (i) KMnO4 – KOH → A; H3O+ → B. (ii) CrO3 → A; H2N–NH–CONH2 → B.

A
NaCN + HCl → A (CN attached to aromatic ring). (C6H5CH2)2Cd + 2CH3COCl → B (Aryl ketone formed).
B
NaCN + HCl → A (Cyanide group added to compound). (C6H5CH2)2Cd + 2CH3COCl → B (Aryl ester formed).
C
NaCN + HCl → A (Aromatic nitrile formed). (C6H5CH2)2Cd + 2CH3COCl → B (Aryl alcohol formed).
D
NaCN + HCl → A (Cyanide attached). (C6H5CH2)2Cd + 2CH3COCl → B (Aryl amide formed).

An element crystallizes in an fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm–3. Calculate the number of atoms in 108 g of the element.

A
The number of atoms in 108 g of the element is 3 × 10^23 atoms.
B
The number of atoms in 108 g of the element is 2 × 10^23 atoms.
C
The number of atoms in 108 g of the element is 4 × 10^23 atoms.
D
The number of atoms in 108 g of the element is 1 × 10^23 atoms.

A 4% solution(w/w) of sucrose (M = 342 g mol–1) in water has a freezing point of 271.15 K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water.

A
The freezing point of the 5% glucose solution is 272.15 K.
B
The freezing point of the 5% glucose solution is 273.15 K.
C
The freezing point of the 5% glucose solution is 270.15 K.
D
The freezing point of the 5% glucose solution is 269.15 K.

When MnO2 is fused with KOH in the presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidizes KI to compound (C), whereas an acidified solution of compound (B) oxidizes KI to compound (D). Identify (A), (B), (C), and (D).

A
(A) MnO2, (B) MnO4–, (C) I2, (D) I–
B
(A) MnO4–, (B) Mn2+, (C) I2, (D) I–
C
(A) Mn2+, (B) MnO4–, (C) I–, (D) I2
D
(A) MnO4–, (B) Mn2+, (C) I2, (D) I2

Write the IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex.

A
IUPAC name: Bis(ethylenediamine)chloroplatinum(II). Geometrical isomers: Cis and Trans.
B
IUPAC name: Chlorobis(ethylenediamine)platinum(II). Geometrical isomers: Cis only.
C
IUPAC name: Ethylenediamine platinum chloride. Geometrical isomers: Trans only.
D
IUPAC name: Platinum chloride ethylenediamine. Geometrical isomers: Cis and Trans.