Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 1 are designed as per the revised syllabus.
CBSE Sample Papers for Class 12 Chemistry Set 1 with Solutions
Time Allowed: 3 Hours
Maximum Marks: 70
General Instructions:
- There are 35 questions in this question paper with internal choices.
- Section A consists of 18 multiple-choice questions carrying 1 mark each.
- Section B consists of 7 very short answer questions carrying 2 marks each.
- Section C consists of 5 short answer questions carrying 3 marks each.
- Section D consists of 2 case-based questions carrying 4 marks each.
- Section E consists of 3 long answer questions carrying 5 marks each.
- All questions are compulsory.
- Use of log tables and calculators are not allowed.
SECTION – A (18 Marks)
(The following questions are multiple-choice questions with one correct answer.
Each question carries 1 mart There is no internal choice in this section.)
Question 1.
The major product of acid catalysed dehydration of 1-methylcyclohexanol is: [1]
(a) 1-methylcyclohexane
(b) 1-methylcyclohexene
(c) 1-cydohexylmethanol
(d) 1-methylenecyclohexane
Answer:
(b) 1-methylcyclohexene
Explanation:
According to Saytzeff rule i.e highly substituted alkene is major product. Here dehydration reaction takes pLace, alkene is formed due to the removal of a water molecule.
Question 2.
Which one of the following compounds is more reactive towards SN1 reaction?[1]
(a) CH2 = CHCH2Br
(b) C6H5CH2Br
(c) C6H5CH (C6H5)Br
(d) C6H5CH(CH3)Br
Answer:
(c) C6H5CH (C6H5)Br
C6H5CH (C6H5) + carbocation formed is more stable
Explanation:
SN1 reaction proceeds by the formation of a carbocation intermediate. Stability of a carbocation is directly proportional to the reactivity to the aLkyl/aryl halide towards SN1.
In C6H5C+(CH3)(C6H5), two phenyl rings diminish the positive charge by their R-effect and CH3 by its + I effect make it stable.
Question 3.
KMnO4 is coloured due to: [1]
(a) d-d transitions
(b) charge transfer from ligand to metal
(c) unpaired electrons in d orbital of Mn
(d) charge transfer from metal to ligand 1
Answer:
(b) charge transfer from ligand to metal
The Mn atom in KMnO4 has +7 oxidation state with electron configuration [Ar]3d °4s° Since no unpaired electrons are present, d-d transitions are not possible. The molecule should, therefore, be colourless.
Its intense purple due to L->M (ligand to metal) charge transfer 2p(L) of O to 3d(M) of Mn.
Question 4.
Which radioactive isotope would have the longer half- life 15O or 19O?
(Given rate constants for 15O or 19O are 5.63 × 10-3 s-1 and k = 2.38 × 10-2 s-1 respectively.) [1]
(a) 150
(b) 190
(c) Both will have the same half-life
(d) None of the above, information given is insufficient
Answer:
(a) 150
The rate constant for the decay of 0-15 is less than that for 0-19 . Therefore, the rate of decay of 0-15 will be slower and will have a longer half life .
Related Theory:
The time in which the concentration of reactants is reduced to half of its initial concentration is called half-life of the reaction. It is denoted as tip.
Question 5.
The molar conductivity of CH3COOH at infinite dilution is 390 Scm2/mol. Using the graph and given information, the molar conductivity of CH3COOK will be: [1]
(a) 100 Scm2/mol
(b) 115 Scm2/mol
(c) 150 Scm2/mol
(d) 125 Scm2/mol
Answer:
(b) 115 Scm2/mol
Λ°CH3COOK = Λ°CH3COOH + Λ°KCl – Λ°HCl
= 390 + 150 – 425 = 115 Scm2/ mol
Question 6.
For the reaction, A + 2B → AB2, the order w.r.t. reactant A is 2 and w.r.t. reactant B. What will be change in rate of reaction if the concentration of A is doubled and B is halved? [1]
(a) increases four times
(b) decreases four times
(c) increases two times
(d) no change
Answer:
(a) increases four times
Rate = [A]2
If [A] is doubled then Rate’ = [2A]2 = 4 [A]2 = 4 Rate
Question 7.
Arrange the following in the increasing order of their boiling points: [1]
A: Butanamine, B: N,N-Dimethylethanamine,
C: N- Etthylethanaminamine
(a) C < B < A
(b) A < B < C
(c) A < C < B
(d) B < C < A
Answer:
(d) B < C < A
In primary amine intermolecular association due. to H-bonding is maximum while in tertiary it is minimum.
Question 8.
The CFSE of [CoCl6]3- is 18000 cm-1 the CFSE for [CoCl4]– will be: [1]
(a) 18000 cm-1
(b) 8000 cm-1
(c) 2000 cm-1
(d) 16000 cm-1
Answer:
(b) 8000 cm-1
Δt = \(\frac{4}{9}\) x 18000cm-1 = 8000 cm-1
Question 9.
What would be the major product of the following reaction? [1]
C6H5 – CH2 – OC6H5 + HBr → A + B
(a) A = C6H5CH2OH, B = C6H6
(b) A = C6H5CH2OH, B = C6H5Br
(c) A = C6H5CH3, B = C6H5Br
(d) A = C6H5CH2Br, B = C6H5OH
Answer:
(d) A = C6H5CH2Br, B = C6H5OH
C6H5CH2OC6H5H + C6H5CH2OC6H5
Explanation:
C6H5 – CH2 – OC6H5 + HBr → C6H5CH6Br + C6H5OH
Caution
Students must understand the mechanism behind every reaction for better identification of major product formed in the reaction.
Question 10.
Which of the following statements is not correct for amines? [1]
(a) Most alkyl amines are more basic than ammonia solution.
(b) pKb value of ethylamine is lower than benzylamine.
(c) CH3NH2 on reaction with nitrous acid releases NO2 gas.
(d) Hinsberg’s reagent reacts with secondary amines to form sulphonamides.
Answer:
(c) CH3NH2 on reaction with nitrous acid releases NO2 gas.
Wrong statement: The evolution of nitrogen gas takes place.
Explanation:
The gas evolved when methylamine reacts with nitrous acid is nitrogen.,
Related Theory
Amines are basic in nature as they react with acids to form salts. As the nitrogen atom of amines possesses a lone pair of electrons, it behaves as Lewis base. The basic strength of amines increases with increase in value of Kb or decrease in the value ofpKb The reactions are shown below.
Question 11.
Which of the following tests/ reactions is given by aldehydes as well as ketones? [1]
(a) Fehling’s test
(b) Tollen’stest
(c) 2,4 DNP test
(d) Cannizzaro reaction
Answer:
(c) 2,4 DNP test
Fehling’s, Tollen’s and Cannizzao reaction is shown by alcohols only.
Question 12.
Arrhenius equation can be represented graphically as follows: [1]
The (i) intercept and (ii) slope of the graph are:
(a) (i) ln A | (ii) Ea/R |
(b) (i) A | (ii) Ea |
(c) (I) ln A | (ii) –Ea/R |
(D) (I) A | (ii) –Ea |
Answer:
(c) (i) In A (ii) – EJR
Explanation:
Arrehnius equation is
lnk = \(\frac{-E_a}{R T}\) + lnA
By comparing the above equation with straight line equation i.e., y = mx + c
lnk = \(\frac{-E_a}{R T}\) + In A
So, m (slope) = \(\frac{-E_a}{R T}\)
C (intercept) = lnA
Question 13.
The number of ions formed on dissolving one molecule of FeS04.(NH4)2S04.6H20 in water is: [1]
(a) 3
(b) 4
(c) 5
(d) 6 1
Answer:
(c) 5
1Fe2+, 2 S042- and 2 NH4+ ions
Explanation:
FeSO4.(NH4)2SO4.6H2O is a double salt.
FeSO4.(NH4)2SO4.6H2O → Fe2+ + 2SO4+ + 2NH+ + 6H2O.
So, 5 ions are formed.
Question 14.
The oxidation of toluene to benzaldehyde by chromyl chloride is called: [1]
(a) Etard reaction
(b) Riemer-Tiemann reaction
(c) Stephen’s reaction
(d) Cannizzaro’s reaction
Answer:
(a) Etard reaction
Explanation:
An end reaction with chromyl chloride forming a precipitated Etard Complex is the first step of this mechanism.
Given below are two statements labelled as Assertion (A) and Reason (R)
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
Question 15.
Assertion (A): An ether is more volatile than an alcohol of comparable molecular mass.
Reason (R): Ethers are polar in nature. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
A and R are two different statements about ethers The correct reason is that hydrogen bonding does not exist amongst ether molecules.
Explanation:
Even though they have the same molecular formula, ethers are more volatile than alcohols. This is due to intermolecular hydrogen bonding in alcohols. H is bonded to the electronegative 0 atom in alcohols. As a result, the H atom of one alcohol’s OH group joins forces with the 0 atom of the second alcohol’s OH group to form a hydrogen bond.
Question 16.
Assertion (A): Proteins are found to have two different types of secondary structures viz alpha-helix and beta-pleated sheet structure. Reason (R): The secondary structure of proteins is stabilized by hydrogen bonding. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
Explanation:
Intrachain H-bonding and minimising steric interference between side chains are two important mechanisms that stabilise the alpha-helix structure.
Caution
It must be noted that proteins are polymers of a-amino acids. Therefore, hydrolysis of proteins yields a-amino acids only.
Question 17.
Assertion (A): Magnetic moment values of actinides are lesser than the theoretically predicted values.
Reason (R): Actinide elements are strongly paramagnetic. [1]
Answer:
(b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
The magnetic moment is less as the 5f electrons of actinides are less effectively shielded which results in quenching of orbital contributions , they are strongly paramagnetic due to presence of unpaired electrons
Question 18.
Assertion (A): Tertiary amines are more basic than corresponding secondary and primary amines in gaseous state.
Reason (R): Tertiary amines have three alkyl groups which cause +1 effect. -1. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation:
On the basis of the +1 group, the basicity order of amines in the gaseous phase can be described. Alkyl groups drive electrons towards nitrogen because they release electrons, which increases the availability of the unshared pair for sharing the basic nature , increases with increase in number of alkyl groups.
SECTION – B (14 Marks)
(The following questions are very short answer type with internal choice in two questions and carry 2 marks each.)
Question 19.
A first-order reaction takes 69.3 min for 50% completion. What is the time needed for 80% of the reaction to get completed?
(Given: log 5 = 0.6990, log 8 = 0.9030, log 2 = 0.3010) [2]
Total Mark | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Answer:
Question 20.
Account for the following:
(A) There are 5 -OH groups in glucose
(B) Glucose is a reducing sugar
OR
What happens when D – glucose is treated with the following reagents? [2]
(A) Bromine water
(B) HNO3
Total Mark | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Answer:
(A) Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five – OH groups. Since it exists as a stable compound, five -OH groups should be attached to different carbon atoms.
(B) Glucose reduces Fehlings reagent
OR
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Question 21.
Give reason for the following:
(A) During the electrophilic substitution reaction of haloarenes, para substituted derivative is the major product.
(B) The product formed during SN1 reaction is a racemic mixture.
OR
(A) Name the suitable alcohol and reagent, from which 2-chloro-2-methyl propane can be prepared.
(B) Out of the chloromethane and fluoromethane , which one is has higher dipole moment and why? [2]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Answer:
(A) At the ortho position, higher steric hindrance is there, hence para isomer is usually predominate and is obtained in the major amount.
(B) During the SN1 mechanism, intermediate carbocation formed is sp2 hybridized and planar in nature. This allows the attack of nucleophile from either side of the plane resulting in a racemic mixture.
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
(A) Tert butyl alcohol or 2-methyl propan- 2-ol using Lucas reagent , mixture of conc. HCl and ZnCl2 the reaction will follow the SN1 pathway.
(B) Chloromethane is having higher dipole moment. Due to smaller size of fluorine the dipole moment of flouromethane is comparatively lesser.
Question 22.
The formula Co(NH3)5CO3Cl could represent a carbonate or a chloride. Write the structures and names of possible isomers. [2]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Answer:
Co(NH3)5CO3Cl and [Co(NH3)5Cl]CO3 Pentaaminecarbonatocobalt(III)chloride Pentaaminechloridocobalt(III)carbonate
Caution
Students often get confused with the IUPAC naming of complexes. The rules must be studied properly in order to name the complexes correctly.
Question 23.
Corrosion is an electrochemical phenomenon. The oxygen in moist air reacts as follows:
O2(g) + 2H2O(l) + 4e– → 4OH–(aq).
Write down the possible reactions for corrosion of zinc occurring at anode, cathode, and overall reaction to form a white layer of zinc hydroxide. [2]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Answer:
Anode: Zn(s) → Zn2+(aq) + 2e–
Cathode: O2(g) + 2H2O(l) + 4e– → 4OH (aq).
Overall: 2 Zn(s) + O2(g) + 2H2O(l) → 2 Zn2+(aq) + 40H–(aq)
2 Zn(S) + O2(g) + 2H2O(l) → 2 Zn(OH)2 (ppt)
Question 24.
Explain how and why will the rate of reaction for a given reaction be affected when: [2]
(A) a catalyst is added
(B) the temperature at which the reaction was taking place is decreased
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Answer:
(A) The rate of reaction will increase. The catalyst decreases the activation energy of the reaction therefore the reaction becomes faster.
(B) The rate of reaction will decrease. At lower temperatures the kinetic energy of molecules decreases thereby the collisions decrease resulting in a lowering of rate of reaction.
Question 25.
Write the reaction and IUPAC name of the product formed when 2-methylpropanal (isobutyraldehyde) is treated with ethyl magnesium bromide followed by hydrolysis. [2]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (VSA) |
|
Answer:
SECTION – C (15 Marks)
(The following questions are short answer type with internal choice in two questions and carry 3 marks each.)
Question 26.
Write the equations for the following reaction: [3]
(A) Salicylic acid is treated with acetic anhydride in the presence of cone. H2SO4
(B) Tert butyl chloride is treated with sodium ethoxide.
(C) Phenol is treated with chloroform in the presence of NaOH
Total Marks | Breakdown (As per CBSE Marking Scheme) |
3m (SA) | Write the reaction in the form of chemical equation in each case. (1m + 1m + 1m) |
Answer:
Question 27.
Using Valence Bond Theory, explain the following in relation to the paramagnetic complex [Mn(CN)6]3-:
(A) Type of hybridization
(B) Magnetic moment value
(C) Type of complex : inner, outer orbital complex: [3]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
3m (SA) |
|
Answer:
[Mn(CN)6]3-
Mn = [Ar] 3d54s2
Mn3+ = [Ar] 3d4
(A) Type of hybridization: d2sp3
(B) Magnetic moment value: \(\sqrt{n(n+2)}\)
= \(\sqrt{(2(2+2))}\) = 2.87 BM
(n = no. of unpaired electrons)
(C) Type of complex: inner orbital
Question 28.
Answer the following questions:
(A) State Henry’s law and explain why are the tanks used by scuba divers filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen)?
(B) Assume that argon exerts a partial pressure of 6 bar. Calculate the solubility of argon gas in water. (Given: Henry’s Law constant for Argon dissolved in water, KH = 40kbar) [3]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
3m (SA) |
|
Answer:
Henry’s law: the partial pressure of the gas in vapour phase (P) is proportional to the mole fraction of the gas (X) in the solution.
The pressure underwater is high, so the solubility of gases in blood increases. When the diver comes to surface the pressure decreases so does the solubility causing bubbles of nitrogen in blood, to avoid this situation and maintain the same partial pressure of nitrogen underwater too, the dilution is done.
Related Theory:
The solubility of gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid. This is called Henry’s law.
(B) P = KHX
Mole fraction of Argon in water X = \(\frac{\mathrm{P}}{\mathrm{K}_{\mathrm{H}}}\)
= \(\frac{6}{40 \times 10^3}\) = 1.5 × 10-4
Question 29.
Give reasons for any 3 of the following observations:
(A) Aniline is acetylated before nitration reaction.
(B) pKfc, of aniline is lower than the m-nitroaniline.
(C) Primary amine on treatment with benzenesulphonyl chloride forms a product which is soluble in NaOH however secondary amine gives product which is insoluble in NaOH.
(D) Aniline does not react with methyl chloride in the presence of anhydrous AlCl3 catalyst. [3]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
3m (SA) |
|
Answer:
(A) Aniline is acetylated, before nitration reaction in order to avoid formation of tarry oxidation products and protecting the amino group, so that p -nitro derivative can be obtained as major product.
(B) pkb of aniline is lower than the m-nitro aniline.The basic strength of aniline is more that m-nitrqaniline. pKb value is inversely proportional to basic strength. Presence of electron withdrawing group decrease basic strength.
(C) Due to the presence of acidic hydrogen in the N-alkylbenzene sulphonamide formed by the treatment of primary amines.
(D) Aniline does not react with methy Ichloride in the presence of AlCl3 catalyst, because aniline is a base and AlCl3 is Lewis acid which lead to formation of salt.
Question 30.
(A) Identify the major product formed when 2-cyclohexylchloroethane undergoes a dehydrohalogenation reaction. Name the reagent which is used to carry out the reaction.
(B) Why are haloalkanes more reactive towards nucleophilic substitution reactions than haloarenes and vinylic halides?
OR
(A) Name the possible alkenes which will yield l-chloro-l-methylcyclohexane on their reaction with HCl. Write the reactions involved.
(B) Allyl chloride is hydrolysed more readily than n-propyl chloride. Why? [3]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
3m (SA) |
|
Answer:
(A) The major product formed when 2-cyclohexylchloroethane undergoes dehydrohalogenation reaction is 1- cyclohexylethene. The reagent which is used to carry out the reaction is ethanolic KOH.
(B) Haloalkanes are more reactive than haloarenes and vinylic halides because of the presence of partial double bond character C-X bond in haloarenes and vinylic halides. Hence they do not undergo nucleophilic reactions easily.
OR
Total Marks | Breakdown (As per CBSE Marking Scheme) |
3m (SA) |
|
(A) Methylenecyclohexane
(B) Allyl chloride shows high reactivity as the carbocation formed in the first step is stabilised by resonance while no such stabilisation of carbocation exists in the case of n-propyl chloride.
SECTION – D (8 Marks)
(The following questions are case-based questions. Each question has an internal choice and carries
4 (1 + 1 + 2) marks, each. Read the passage carefully and answer the questions that follow.)
Question 31.
Strengthening the Foundation: Chargaff Formulates His “Rules”
Many people believe that James Watson and Francis Crick discovered DNA in the 1950s. In reality, this is not the case. Rather, DNA was first identified in the late 1860s by Swiss chemist Friedrich Miescher. Then, in the decades following Miescher’s discovery, other scientists—notably, Phoebus Levene and Erwin Chargaff—carried out a series of research efforts that revealed additional details about the DNA molecule, including its primary chemical components and the ways in which they joined with one another. Without the scientific foundation provided by these pioneers, Watson and Crick may never have reached their groundbreaking conclusion of 1953: that the DNA molecule exists in the form of a three-dimensional double helix.
Chargaff, an Austrian biochemist, as his first step in this DNA research, set out to see whether there were any differences in DNA among different species.
After developing a new paper chromatography method for separating and identifying small amounts of organic material, Chargaff reached two major conclusions:
(I) The nucleotide composition of DNA varies among species.
(II) Almost all DNA, no matter what organism or tissue type it comes from maintains certain properties, even as its composition varies. In particular, the amount of adenine (A) is similar to the amount of thymine (T), and the amount of guanine (G) approximates the amount of cytosine
(C) In other words, the total amount of purines (A + G) and the total amount of pyrimidines (C + T) are usually nearly equal. This conclusion is now known as “Chargaff s rule.”
Chargaff’s rule is not obeyed in some viruses. These either have single- stranded DNA or RNA as their genetic material.
(A) A segment of DNA has 100 adenine and 150 cytosine bases. What is the total number of nucleotides present in this segment of DNA? [1]
(B) A sample of hair and blood was found at two sites. Scientists claim that the samples belong to same species. How did the scientists arrive at this conclusion? [1]
(C) The sample of a virus was tested and it was found to contain 20% adenine, 20% thymine, 20 % guanine and the rest cyto-sine, Is the genetic material of this virus (a) DNA- double helix (b) DNA-single helix (c) RNA? What do you infer from this data?
OR
How can Chargaff’s rule be used to infer that the genetic material of an organism is double- helix or single- helix? [2]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
4m (CBQ) |
|
Answer:
(A) A = 100 and T = 100 C = 150 and G = 150
Total nucleotides = 100 + 100 + 150 + 150 = 500
(B) They studied the nucleotide composition of DNA. It was the same so they concluded that the sampLes belong to same species.
(C) A = T = 20%
But G is not equal to C so double helix is ruled out.
The bases pairs are ATGC and not AUGC so it is not RNA
The virus is a single helix DNA virus
OR
Total Marks | Breakdown (As per CBSE Marking Scheme) |
2m (CBQ) | (C) Mention the inference of the gentic material by Chargaffs rule. (2m) |
According to Charagraff rule, all double helix DNA will have the same amount of A and T as well as C will be same amount as G. If this is not the case then the helix is single stranded.
Question 32.
Henna is investigating the melting point of different salt solutions.
She makes a salt solution using 10 ml of water with a known mass of NaCl salt.
She puts the salt solution into a freezer and leaves it to freeze.
She takes the frozen salt solution out of the freezer and measures the temperature when the frozen salt solution melts.
She repeats each experiment.
Assuming the melting point of pure water as 0°C, answer the following questions:
(A) One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer. [1]
(B) Why did Henna collect two sets of results? [1]
(C) In place of NaCE, if Henna had used glucose, what wouLd have been the melting point of the solution with 0.6 g glucose in it?
OR
What is the predicted melting point 1.2 g of salt is added to 10 ml of water? Justify your answer. [2]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
4m (CBQ) |
|
Answer:
The melting point of ice is the freezing point of water. We can use the depression in freezing point property in this case.
(A) 3rd reading for 0.5 g there has to be an increase in depression of freezing point and therefore decrease in freezing point so also decrease in melting point when amount of salt is increased but the trend is not followed on this case.
(B) Two sets of reading help to avoid error in data collection and give more objective data.
(C)
Freezing point or Melting point = – 0.62 °C
OR
Total Marks | Breakdown (As per CBSE Marking Scheme) |
4m (CBQ) |
|
Depression in freezing point is directly proportional to molality (mass of solute when the amount of solvent remains same)
0.3 g depression is 1.9 °C
0.6 g depression is 3.8 °C
1.2 g depression will be 3.8 × 2 = 7.6 °C
SECTION – E (15 Marks)
(The following questions are long answer type and carry 5 marks each.
Two questions have an internal choice.)
Question 33.
(A) Why does the cell voltage of a mercury cell remain constant during its lifetime?
(B) Write the reaction occurring at anode and cathode and the products of electrolysis of aq. KCl.
(C) What is the pH of HCl solution when the hydrogen gas electrode shows a potential, of -0.59 V at standard temperature and pressure?
OR
(A) Molar conductivity of substance “A” is 5.9 × 103 S/m and “B” is 1 × 10-16 S/m. Which of the two is most likely to be copper metal and why?
(B) What is the quantity of electricity in Coulombs required to produce 4.8 g of Mg from molten MgCl2? How much Ca will be produced if the same amount of electricity was passed through molten CaCl2? (Atomic mass of Mg = 24 u, atomic mass of Ca = 40 u).
(C) What is the standard free energy change for the following reaction at room temperature? Is the reaction spontaneous?
Sn(s) + 2Cu2(aq) → Sn2+(aq) 2Cu(s) [5]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
5m (LA) |
|
Answer:
(A) The cell potential remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life time.
(B) KCl(aq) → K+(aq) + Cl–(aq)
Cathode: H2O(l) + e– → ½ H2(g) + OH_(aq)
Anode: Cl–(aq) → ½ Cl2(aq) + e–
net reaction:
KCl(0q) + H2O(l) → K+(aq) + OH (aq) + ½ H2(g) + ½Cl2(g)
(C) Given, potential of hydrogen gas electrode = -0.59 V
Electrode reaction: H+ + e– → 0.5 H2
Applying Nernst equation,
n = 1
[H2] = 1 bar
-0.59 = 0 – 0.059 ( – log [H+])
-0.59 = -0.059pH
∴ pH = 10
OR
Total Marks | Breakdown (As per CBSE Marking Scheme) |
5m (LA) |
|
(A) “A” is copper, metals are conductors thus have high vaLue of conductivity.
(B) Mg2+ + 2e– → Mg
1 mole of magnesium ions gains two moles of electrons or 2F to form 1 mole of Mg
24 g Mg requires 2 F electricity 4.8 g Mg requires 2 × 4.8/24 = 0.4 F = 0.4 × 96500 = 38600C
Ca2+ + 2e– → Ca
2 F electricity is required to produce 1 mole = 40 g Ca
0.4 F electricity will produce 8 g Ca
(C) F = 96500C, n = 2,
Sn2+(aq) + 2e– → Sn(s) E° = 0.14V
Cu2+(aq) + e– → Cu+(aq) E° = 0.15 V
E°cell = E°cathode – E°anode
= 0.15 – (-0.14) = 0.29V
ΔG° = -nFE°cell
= -2 × 96500 × 0.29 = 55970 J/mol
Question 34.
A hydrocarbon (a) with molecular formula C5H10 on ozonolysis gives two products (b) and (c). Both (b) and (c) give a yellow precipitate when heated with iodine in presence of NaOH while only (b) give a silver mirror on reaction with Tollen’s reagent.
(A) Identify (a), (b) and (c).
(B) Write the reaction of (b) with Tollen’s reagent
(C) Write the equation for iodoform test for (c)
(D) Write down the equation for aldol condensation reaction of (b) and (c).
OR
An organic compound (a) with molecular formula C2Cl3O2H is obtained when (b) reacts with Red P and Cl2. The organic compound (b) can be obtained on the reaction of methyl magnesium chloride with dry ice followed by acid hydrolysis.
(A) Identify (a) and (b)
(B) Write down the reaction for the formation of (a) from (b). What is this reaction called?
(C) Give any one method by which organic compound (b) can be prepared from its corresponding acid chloride.
(D) Which will be the more acidic compound (a) or (b)? Why?
(E) Write down the reaction to prepare methane from the compound (b). [5]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
5m (LA) |
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Answer:
(A) (a) is an alkene CH(CH3) = C(CH3)2
(b) is an aldehyde with -CH3 group CH3CHO
(c) is a methyl ketone
(B) CH3CHO + [Ag(NH3)2]+ + OH– → CH3COO– + Ag + NH3 + H2O
(C) CH3COCH3 + NaOH + I2 → CHI3 + CH3COONa
O = C(CH3)2
(D) CH3COCH3 + CH3CHO
OR
Total Marks | Breakdown (As per CBSE Marking Scheme) |
5m (LA) |
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(A) (a): CCl3COOH (b) : CH3COOH
(D) a will be more acidic due to presence of 3 Cl groups (electron withdrawing groups) which increase acidity of carboxylic acid.
Question 35.
Answer the following:
(A) Why are all copper halides known except that copper iodide?
(B) Why is the E°(V3+/V2+) value for vanadium comparatively low?
(C) Why HCl should not be used for potassium permanganate titrations?
(D) Explain the observation, at the end of each period, there is a slight increase in the atomic radius of d-block elements.
(E) What is the effect of pH on dichromate ion solution? [5]
Total Marks | Breakdown (As per CBSE Marking Scheme) |
5m (LA) |
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Answer:
(A) Cu2+ oxidizes iodide ion to iodine.
(B) The low value for V is related to the stability of V2+ (half-filled t2g level)
(C) Permanganate titrations in presence of hydrochloric acid are unsatisfactory since hydrochloric acid is oxidised to chlorine.
(D) The d-orbital is full with ten electrons and shield the electrons present in the higher s-orbital to a greater extent resulting in increase in size.
(E) The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. Increasing the pH (in basic solution) of dichromate ions a colour change from orange to yellow is observed as dichromate ions change to chromate ions.