CBSE Sample Papers for Class 12 Chemistry Paper 5 are part of CBSE Sample Papers for Class 12 Chemistry. Here we have given CBSE Sample Papers for Class 12 Chemistry Paper 5.
CBSE Sample Papers for Class 12 Chemistry Paper 5
Board | CBSE |
Class | XII |
Subject | Chemistry |
Sample Paper Set | Paper 5 |
Category | CBSE Sample Papers |
Students who are going to appear for CBSE Class 12 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 5 of Solved CBSE Sample Paper for Class 12 Chemistry is given below with free PDF download solutions.
Time Allowed : 3 Hours
Max. Marks : 70
General Instructions
- All questions are compulsory.
- Question number 1 to 5 are very short answer questions and carry 1 mark each.
- Question number 6 to 10 are short answer questions and carry 2 marks each.
- Question number 11 to 22 are also short answer questions and carry 3 marks each.
- Question number 23 is a value based question and carry 4 marks.
- Question number 24 to 26 are long answer questions and carry 5 marks each.
- Use log table, if necessary. Use of calculators is not allowed.
Questions
Question 1.
Which type of defect is present in metallic hydride of transition metals?
Question 2.
What is electrophoresis?
Question 3.
Give an example of a chelating ligand.
Question 4.
Write IUPAC name of the following:
Question 5.
Arrange the following in increasing order of pKb values:
p-nitroaniline, Aniline, p-Toluidine
Question 6.
Describe the preparation of kMnO4. How does the acidified permanganate solution react with oxalic acid? Write the ionic equations for the reactions.
OR
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with
(i) iodine,
(ii) H2S.
Question 7.
Write the mechanism of ether formation from ethanol by its reaction with conc. H2SO4 at 410 K.
Question 8.
Define the following terms
- Reverse osmosis
- Azeotropic solution
Question 9.
For an elementary reaction
2A + B → 3C
the rate of appearance of ‘C’ at time T is 1.3 × 10-4 mol L-1 s-1. Calculate at this time the
(i) rate of reaction
(ii) rate of disappearance of A
Question 10.
Explain the following:
- White phosphorous is more reactive than red phosphorous
- NF3 is an exothermic compound but NCl3 is an endothermic compound.
Question 11.
A, 5% solution (by mass) of cane sugar (m.wt 342) is isotonic with 0.877% solution of a substance X. Find the molecular weight of X (consider density of both solution as 1 g/cm3).
Question 12.
Describe the following:
- Electrodialysis
- Cotrell smoke precipitator
- Hardening of leather
Question 13.
Answer the following:
- In Hall heroult process of extraction of Al anode must be of graphite only. Give reason.
- Give differences between roasting and calcination.
- Explain autoxidation with a suitable example.
Question 14.
The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce
the initial concentration to its \(\frac { 1 }{ 10 }\) th value?
Question 15.
Silver crystallises in fee arrangement having radii 141.4 pm and its density is 8 g/cm3. Find the number of unit cells in its 10 cm3 volume.
OR
Explain the following terms:
- Ferromagnetic solids
- p-type semiconductor
- F-centre
Question 16.
What is meant by disproportionation? Give one example of disproportionation reaction in aqueous solution.
Question 17.
Arrange the following in increasing order as per the given instruction:
- 1-Bromobutane, 2-Bromobutane, 2-Bromo-2-methyl propane (Tendency for SN2 reaction)
- CH3F, CH3Br, CH3Cl (Boiling point)
- CHCl3, CCl4, CH2Cl2 (Dipole moment)
Question 18.
Carry out following conversions
(a) But-1-ene to 1-Butanol
(b) Ethyl magnesium chloride to propan-1-ol
(c) Benzyl chloride to 2-phenyl ethanoic acid
Question 19.
- Write IUPAC name of the following
K4[Fe(CN)6] - Write the formula for hydrogen tetrafluorido borate (III)
- Write the linkage isomer of [Co(NH3)5NO2]2+
Question 20.
Write the major product in the following reaction:
(i) CH3CH2OH \(\underrightarrow { { PCl }_{ 5 } }\)
(iii) CH3Cl + CH3CH2-ONa →
Question 21.
Define the following
- Peptide linkage
- Hormones
- Enzymes
Question 22.
Explain with example
- Addition copolymer
- Condensation homopolymer
- Biodegradable polyester
Question 23.
Neeraj went to the departmental stone to purchase grocery items. On one of the shelves he noticed sugar free tablets. He decided to buy then for his grandfather who was a diabetic. There were three types of sugar free tablets. He decided to buy sucralose which was good for his grandfather’s health.
- Name another sugar free tablet which Neeraj did not purchase.
- Was it right to purchase such medicines without doctor’s prescription?
- What quality of Neeraj is reflected the above act?
Question 24.
(a) Define conductivity and molar conductivity for the solution of an electrolyte. Discuss variation of these with concentration.
(b) Calculate the standard cell potential of the galvanic cell in which the following reaction takes place:
Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Also calculate ΔG° and equilibrium constant of the reaction.
OR
- Why on dilution the Δm of CH3COOH increases drastically while that of CH3COONa increases gradually.
- Write the mathematical expression for kohlrausch’s law.
- Which type of metals can be used in cathodic protection of iron against rusting?
- What are secondary cells?
Question 25.
- Which poisonous gas is evolved when white phosphorous is heated with cone. NaOH solution? Write the chemical equation.
- Write the formula of first noble gas compound prepared by N-Bartlett. What inspired him to prepare this compound.
- Complete the following equation
CaF2 + H2SO4 →
OR
(a) Assign reasons for the following
- Bi (V) is a stronger oxidising agent than Sb(V).
- Of all the noble gases only Xenon is known to form established compounds.
(b) Draw structures of the following molecules:
(i) H2S2O6
(ii) (HPO3)3
(iii) ClF5
Question 26.
(a) Draw the structures of the following:
(i) p-methyl benzaldehyde
(ii) 4-methyl pent-3-en-2-one
(b) Give chemical tests to distinguish between the following pairs of compounds:
- Benzoic acid and ethyl benzoate
- Acetophenone and benzophenone
- Phenol and benzaldehyde
OR
(a) Draw the structures of the following derivatives
(i) 2,4-DNP of Cyclohexanone
(ii) Oxime of Acetone
(b) How will you convert the following
(i) Benzaldehyde to toluene
(ii) Benzyaldehyde to 3-Hydroxy-3-phenyl propanol
(iii) Benzaldehyde to Acetophenone
Answers
Answer 1.
Interstitial defect
Answer 2.
Electrophoresis: The movement of colloidal particles towards oppositely charged electrode in an electric field is called electrophoresis.
Answer 3.
Ethane-1, 2-diol, Ehane-1, 2-diamine, EDTA (anyone).
Answer 4.
3-Phenyl prop-2-enal.
Answer 5.
p-nitroaniline < aniline < p-Toluidine
Answer 6.
Preparation of KMnO4:
Overall reaction →
2MnO–4 + 16H+ + 5C2O2-4 → 2Mn2+ + 8H2O + 10CO2
OR
Potassium dichromate acts as an oxidising agent in acidic medium and gets reduced to Cr3+ ion.
Answer 7.
Mechanism of “Etherification”
Answer 8.
- Reverse osmosis: It is the movement of solvent molecules from higher concentration solution to lower concentration solution through a semipermeable membrane, when external pressure is applied on higher concentration solution side of membrane. (It is used for the purification of water).
- Azeotropic solution: A binary solution which has the same composition in liquid state as well as in vapour state and cannot be separated by fractional distillation process is known as azeotropic solution.
Answer 9.
For the reaction
2A + B → 3C
Answer 10.
- P4 exists as discrete tetrahedral units in which there is high ring strain as compared to the polymeric red phosphorous, hence white phosphorous is more reactive.
- Since bond enthalpy of F2 is much smaller than Cl2 due to lone pair-lone pair repulsions and N – F bond is stronger than N – Cl bond. As a result NF3 formation is an exothermic reaction while NCl3 formation is an endothermic reaction.
Answer 11.
For 5% solution of cane sugar, π = CRT
Answer 12.
- Electrodialysis: It is the process of removing a dissolved electrolytic substance from a colloidal solution by applying electric field, which separates out the electrolyte dissolved from the non-electrolytic colloidal solution by dialysis process.
- Cottrell smoke precipitator: Smoke is a colloidal solution consisting of different particles like dust, metals, arsenic and other harmful gases. On passing smoke containing air through a chimney which leads to a chamber consisting of charged particles opposite to that of smoke particles, precipitates out the smoke particles from the air and in this way air is cleaned. Such precipitator is called cottrell smoke precipitator.
- Hardening of leather: When a hide containing positively charged particles is soaked in tannin, which has negatively charged particles it results in coagulation of particles and thus hardening of leather takes place. This process is called tanning.
Answer 13.
- At anode, O2 is liberated which combines with graphite electrode and form CO2 which
covers the surface of molten aluminium formed during the process, otherwise aluminium will again get oxidised by the released O2 at anode. - Roasting is a process in which ore is heated below its melting point in sufficient supply of O2 to eliminate the volatile oxides (SO2, P2O5). On the other hand in calcination, ore is heated below its melting point in the absence of air, so that volatile impurities like CO2, moisture, organic matter can be expelled.
- Auto-oxidation: The process in which no oxidising agent needs to be added to get the metal is known as auto-oxidation, e.g. during the extraction of copper no oxidising or reducing agent is required. The reaction involved is
- Cu2O + Cu2S → Cu + SO2
S-2 get oxidised to SO2 and Cu2+ gets reduced to copper. The oxide ion already present oxidises S2- ion to SO2.
Answer 14.
Answer 15.
For fcc arrangement Z = 4 and
OR
- Ferromagnetic solid: The solids in which all the molecular domains get arranged in the same direction permanently under the influence of electrical or magnetic field are known as ferromagnetic solids. Such solids can be converted in permanent magnets and are strongly attracted towards the magnetic field.
- p-type semiconductor: When an element (like silicon) doped with other element having less valence electrons (like Boron), electron vacancy or holes are generated in the element lattice and are this is responsible for electrical conduction. Such a conductor is called p-type semiconductor.
- F-centre: In metal excess defect, electrons get trapped in the anionic vacancies and such positions are known as F-centres, which is also responsible for the appearance of colour in the crystal.
Question 16.
Disproportionation: The reaction in which same substance get oxidised and reduced simultaneously is known as disproportionation reaction.
Example:
- Dissociation Cu+ in aqueous solution
Cu+ + H2O → Cu2+ (aq) + Cu (s) - \(\overset { 0 }{ { Cl }_{ 2 } } \)(g) + H2O(l) → HCl-1 + HOCl+1
Answer 17.
- 2-Bromo-2-Methyl propane < 2-Bromobutane < 1-Bromobutane
- CH3F < CH3Cl < CH3Br
- CCl4 < CHCl3 < CH2Cl2
Answer 18.
Answer 19.
- Potassium hexayanidoferrate (II)
- H[BF4]
- [CO(NH3)5(ONO)]2+
Answer 20.
(a) CH3CH2OH \(\underrightarrow { { PCl }_{ 5 } }\) CH3CH2Cl
(c) CH3Cl + CH3CH2ONa → CH3CH2-O-CH3
Answer 21.
- Peptide linkage: The
linkage, i.e. amide linkage present between the two a-amino acids is known as peptide linkage in protein chemistry. - Hormones: They are the biochemical substances produced by ductless glands (endocrine glands) such as thyroid, adrenal, etc. which acts on specific organs to regulate the functions of body organ.
- Enzymes: These are proteins which acts as biological catalysts e.g. maltase, lactase, invertase etc and acts at normal temperature and pressure.
Answer 22.
1. Addition copolymer: The polymerisation reaction which takes place in unsaturated compounds without the loss of any small molecule and have two or more type of monomers forms a polymer known as addition copolymer, e.g. Buna-N, Buna-S etc.
2.Condensation homopolymer: The momomer having multifunctional group undergo polymerisation reaction with loss of small molecules like H2O, NH3 etc. to form a polymer are known as condensation homopolymer, e.g. Nylon-6.
3. Biodegradable polyester: The polymer which undergoes bacterial degradation in the environment are known as biodegradable polymer, e.g. PHBV – Polyhydroxybutyrate – co – β-hydroxy valerate. Used for speciality packaging, orthopaedic devices, controlled drug release etc.
Answer 23.
- Sacharine, Aspartame
- No, Doctor is expert for this.
- Concern for grandparents, caring attitude, critical thinking
Answer 24.
(a) Conductivity: The ability of a solution to conduct electric current is called conductivity.
K = \(\quad \frac { { Cell\quad constant\quad G }^{ \ast } }{ R } \)
Molar conductivity: The molar conductivity is defined as the conductivity of an electrolytic solution containing 1 mole of the solute between two parallel electrodes separated by 1 cm.
Variation with concentration: Conductivity decreases with dilution, the reason being is the no. of ions per unit volume decreases. On the other hand molar conductivity will increase due to increase in volume of the solution for the same one mole of electrolyte.
(b) Fe2+ → Fe3+ + e Oxidation half reaction
Ag+ + e → Ag Reduction half reaction
Change in number of electrons → n = 1
OR
- Since acetic acid is a weak electrolyte therefore after a particular extent of dilution its ionisation will increase abruptly. Hence there is a drastic increase in its molar conductivity. On the other hand CH3COONa is salt and it is always 100% ionised in its aqueous solution. Hence on dilution only interionic attraction forces decreases, so there is only a minor increase in its molar conductivity.
- If AxBy is an ionic compound than
- More active metals than Fe, e.g. Zn, Mg etc. having more negative reduction potential than Fe.
- The electrochemical cells which can be recharged and used again after once used, for a longer duration are known as secondary cells.
Answer 25.
- PH3 (phosphine) a poisonous gas is formed. The reaction is
- P4 + NaOH → PH3 + NaH2PO2
- Xe [PtF6]
Formation of compound of molecular O2 wilts PtF6, inspite of having ionisation energy of O2 as 1175 kJ/mol similar to the ionisation energy of 1170 kJ/mol inspired Bartlett to form compounds of Xe, i.e. first noble gas compound. - CaF2 + H2SO4 → CaSO4 + 2HF
OR
(a)
- Due to inert pair effect, the lower oxidation state, two less than the group oxidation state becomes more stable on moving down the group and as a result Bi(III) is more stable than Bi(V). Hence Bi(V) is a stronger oxidising agent than Sb(V).
- Xenon is of larger size and have the lowest ionisation energy among all the noble gases, hence it is able to form compounds with highly electronegative atoms like oxygen and fluorine.
Answer 26.
(b) Distinguish
1.
Reagent |
Benzoic acid |
Etthyl benzoate |
NaHCO3(aq) |
Brisk effervesence of CO2 |
No gas evolves |
2.
Reagent |
Aletophenone |
Benzophenone |
NaOH + I2 + heat | CHI3, yellow ppt. forms | No yellow ppt. formation |
3.
Reagent |
Phenol |
Benzoldehyde |
Neutral FeCl3 |
Dark violet grey colour forms |
No colour change |
OR
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