CBSE Sample Papers for Class 12 Chemistry Set 7 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Chemistry with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Chemistry Set 7 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.

follows: [1]
(a) Raoult’s law
(b) negative deviation from Raoult’s law
(c) positive deviation from Raoult’s law
(d) Henry’s law at all compositions
Answer:
(c) positive deviation from Raoult’s Law

Explanation:
A solution shows positive deviation from Raoult’s law any composition, its vapour presseue is higher than that given by Raoult’s law.

Related Theory:
Solutions with weaker A-B interactions than A-A or B-B interactions demonstrate a positive divergence from Raoult’s rule. These mixtures produce minimum boiling azeotropes. Solutions where the A-B interactions are stronger than the A-A or B-B interactions demonstrate a negative deviation from Raoult’s law. These mixtures provide the highest boiling azeotropes.

Question 2.
Which of the following is not the reducing sugar from the following? [1]
(a) Fructose
(b) Galactose
(c) Sucrose
(d) Lactose
Answer:
(c) Sucrose

Explanation:
All monosaccharides and disaccrides except sucrose are reducing sugars. Fructose and galactose are monosaccharides and lactose is a dissachrides.

Related Theory:
Non-reducing sugars are the carbohydrates which do not contain free functional group and so they do not reduce Fehling’s or Tollen’s reagent. All polysaccharides are non reducing.

Question 3.
With the help of the graph given below answer the following question.

Why there are gradual decrease in ionic radii in going from Ti⁺ ion to Cu⁺
(a) Due to decrease in effective nuclear charge.
(b) Due to increase in effective nuclear charge.
(c) Due to increase in ionic potential.
(d) Due to decrease in ionic potential.
Answer:
(b) Due to increase in effective nuclear charge.

Explanation:
The nuclear charge gradually grows from left to right over time. The electron enters the same sheLl as the atomic number rises over time. They get increasingly drawn to the nucleus as a result. As a result of this force of attraction from the nucleus, the ionic radii gradually decreases.

Question 4.
Which of the following is the condition for solutions showing positive deviation in case of non-ideal solutions? [1]
(I) P = P_A + P_B = P°_AX_A + P°_BX_B
i.e.,
P_A = P°_A x_A; P_B = P°_BX_B
(II) P_A > P°_AX_A; P_B > P°_BX_B
(III) P_A < P°_AX_A; P_B < P°_BX_B
(a) Only (I)
(b) Only (II)
(c) Both (I) and (III)
(d) All of the above
Answer:
(b) Only (II)

Explanation:
A solution is said Lo show positive deviation from Raoult’s law at any composition, its vapour pressure is higher than that given by Raoult’s Law.

Question 5.
The common name of

is: [1]
(a) N-Diethyl benzyl
(b) N-Bimethyl aniline
(c) N, N-Dimethyl-aniline
(d) 2° amine
Answer:
(c) N, N-Dimethyl-aniline

Explanation:
As two methyl groups are attached to -N, so it is named as N,N-dimethyl aniline.

Question 6.
The octahedral complexes are formed either by d²sp³ or sp³d² hybridisation and can be grouped into two categories i.e. outer and inner orbital complex.
Which of the following is/are the example of outer orbital complex? [1]
(a) [COF₆]^3-
(b) Zn(NH₃)₆]²⁺
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Explanation:
Complexes which involves the sp³d² hybridisation are called outer orbital complexes and both [CoF₆]^3- and Zn(NH₃)₆]²⁺ are the examples of outer orbital complexes. Because they involves the participation of outer d-orbitals.

Question 7.
Write the odd one out from the following coordination complexes. [1]
[Hgl₃]^–, [ZnCl₄]^2-, [Ni(CO)₄], [MnCl₄]^2-
(a) [Hgl₃]^–
(b) [ZnCl₄]²
(c) [Ni(CO)₄]
(d) [MnCl₄]^2-
Answer:
(a) [Hgl₃]^–

Explanation:
[Hgl₃]^– have the geometry of trigonal planar while rest of the coordination complexes shows the geometry of tetrahedral.

Question 8.
R-NO₂ + 6H → RNH₂ + 2H₂O [1]
Write the missing reagent used in above reaction.
(a) HNO₃
(b) Sn/HCl
(c) H₂O₂
(d) K₂NO₃
Answer:
(b) Sn/HCl

Explanation:
This is the reaction of reduction of nitro compound which is done in the presence of reducing agents. So, Sn/HCl is a reducing agent while other options are the oxidizing agents.

Question 9.
Order of dehydration of alcohol follows: [1]
(a) 2° alcohol > 3° alcohol > 1° alcohol
(b) 3° alcohol > 2° alcohol > 1° alcohol
(c) 1° alcohol > 3° alcohol > 2° alcohol
(d) 3° alcohol < 2° alcohol > 1° alcohol
Answer:
(b) 3° alcohol > 2° alcohol > 1° alcohol

Explanation:
the ease of formation of alkene i.e. the ease of dehydration of alcohol depends upon the stability of carbocation formed during dehydration of alcohol. Since, the stability of carboctaions follows the order 3° > 2° > 1°, the ease of dehydration also follows the same order.

Question 10.
Glucose is a/an: [1]
(a) aldopentose
(b) ketohexose
(c) aldohexose
(d) ketopentose
Answer:
(c) aldohexose

Explanation:
Glucose (also known as dextrose) is a carbohydrate compound consisting of six carbon atoms and an aldehyde group and they are referred to as aldohexose.

Question 11.
The term used for a solution that has higher osmotic pressure than its standard solution is: [1]
(a) Hypotonic
(b) Hypertonic
(c) Isotonic
(d) Concentrated
Answer:
(b) Hypertonic

Explanation:
A solution having higher osmotic pressure than its standard solution is said to be called hypertonic solution. A solution having a lower osmotic pressure relative to its standard solution is called hypotonic solution. Solutions which have the same osmotic pressure as that of its standard solutions are termed as isotonic or iso-osmotic solutions.

Question 12.
The plant cell will shrink when placed in: [1]
(a) water
(b) a hypotonic solution
(c) a hypertonic solution
(d) an isotonic solution
Answer:
(c) a hypertonic solution

Explanation:
Hypertonic solutions have higher osmotic pressure and therefore they have higher concentration. When a plant cell is kept in hypertonic solution, plant cell loses water and therefore the plant cell gets shrinked.

Question 13.
Which is the correct structure for the following compound? [1]

(a) 1-Bromo-2-chloro-4-methylhexane
(b) 1-Bromo-2-chloro-4-methylpentane
(c) 1-Bromo-3-chloro-4-methylpentane
(d) 5-Bromo-4-chloro-4-methylpentane
Answer:
(b) 1 -Bromo-2-chloro-4-methylpenane

Explanation:
Alphabetically, the halogens are arranged bromo at position 1 and chloro at position 2 and word root for this unsaturated compound is pentane.

Question 14.
Give the IUPAC name of the following:
CH₃CH(Cl)CH(Br)CH₃
(a) 3-Bromo-2-chloro butane
(b) 3-Bromo-3-chloro butane
(c) 2-Bromo-2-chloro butane
(d) 2-Bromo-3-chloro butane [1]
Answer:
(d) 2-Bromo-3-chioro butane

Explanation:
Alphabetically the bromo derivative is named first followed by chloro and word root is butane as it is a four carbon atom saturated chain.

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(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: Alkylamines undergo electrophilic substitution reaction.
Reason: NH₂ group present on the benzene ring behaves as an o- and p- directing group and directs the incoming electrophile at the ortho and para-position of the aromating ring. [1]
Answer:
(c) (A) is true but (R) is false

Explanation:
Arylamines like aniline shows
the mechanism of electrophilic substitution reaction at ortho and para positions of the ring as NH₂ group present on the benzene ring behaves as an o- and p- directing group and directs the incoming electrophile at the ortho and para-position of the aromating ring.

Question 16.
Assertion: Slowest elementary step is also refereed as rate determining step.
Reason: The rate of the reaction is determined by the rate of slowest elementary step. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Since the rate of the slowest elementary step determines the pace of the reaction, it is also known as the rate determining step.

Question 17.
Assertion: Acetaldehyde and formaldehyde can be distinguished by iodoform test
Reason: Acetaldehyde gives a yellow ppt when heated with iodine while formaldehyde does not give this test. [1]
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).

Explanation:
Acetaldehyde and formaldehyde can be distinguished by iodoform test. Acetaldehyde gives a yellow ppt when heated with iodine while formaldehyde does not give this test.

Question 18.
Assertion: Phenol develop red or pink color on long standing.
Reason: Phenol turns pink in colour on exposure to air due to slow reduction. [1]
Answer:
(d) (A) is false but (R) is true

Explanation:
When phenol is exposed to air, it slowly oxidises and becomes pink. Phenoquinone, which is pink in colour, is created when quinone, which was first created from phenol, is combined with phenol.

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.
Write the oxidation number of:
(A) Fe in K₃[Fe(CN)₆]
(B) Au in H[AuCl₄]
OR
What type of isomerism is shown by the following compounds? [2]
(A) [Cr(SCN)(H₂O)₅]²⁺ and [Cr(NCS(H₂O)₅]²⁺
(B) [Cu(NH₃)₄][PtCl₄] and [Pt(NH₃)₄][CuCl4]
Answer:
(A) +3 + X + (-1) × 6 = 0 X = +3
(B) +1 + X+ (-1) × 4 = 1 X = +3
OR
(A) Linkage isomerism
(B) Coordination isomerism

Question 20.
Elaborate the composition and role of these following reagents in reactions: [2]
(A) Fehling solution
(B) Tollen’s reagent
Answer:
(A) Fehling’s solution, an intensely blue alkaline solution, is used to detect the presence of aldehydes or groups containing any aldehyde functional group (-CHO), as well as to distinguish between reducing and non-reducing sugars using Tollen’s reagent. To distinguish between a ketone group and water-soluble carbohydrates, utilise Fehling’s solution.

(B) It is a ammonical solution of silver nitrate. It is formed by the addition of dilute solution of ammonium hydroxide to silver nitrate till the formed dirty white ppt of AgOH is dissolved.

Question 21.
Write the equation of the formation of following compounds showing nucleophilic substitution reaction.
(A) CH₃CH₂CH₂CN
(B) CH₃CH₂CH₂CH₂NO₂
OR
How will you carry out the conversion of chlorobenzene to:
(A) Salicyladehyde
(B) Phenol [2]
Answer:

OR

Question 22.
Compute the graphical behaviour of first order reaction by plotting on graph:
(A) concentration against time
(B) log_1o[A] against time [2]
Answer:
(A)

(B)

Related Theory:
The plot of the logarithm of [A] versus time is a straight line with k = – slope of the line.

Question 23.
(A) For a reaction A + B → P, the rate law is given by, r = k [A]^1/2 [B]². What is the order of this reaction?
(B) A first-order reaction is found to have a rate constant k = 5.5 × 10^-1414 s^-1. Find the half-life of the reaction. [2]
Answer:
(A) \(\frac{5}{2}\)

(B) t_½ = 1.26 × 10¹³ s
(Hint: r = k[A]^½ [B]², t^½ = 0.693/k)

Caution:
You can judge the reaction order by knowing the unit of rate constant General formula for units of rate constant is found out using the following formula

Question 24.
Write down the deficiency disease and chemical names of following vitamins. [2]
(A) Vitamin B₁
(B) Vitamin K
Answer:
(A) Disease: beri-beri , chemical name: thiamine
(B) Disease: haemorrhage, chemical name: phylloquinone

Question 25.
An alcohol ’A’ of molecular formula (C₂H₆O) carried out with cone. Solution of HI and gives ‘B’. ‘A’ is treated with HCl in the presence of dehydrating agent gives product ‘C’. Find out ‘A’, ‘B’ and ‘C’. [2]
Answer:

Section – C (15 Marks)
(The following questions are short answer type with internal choice in one question and carry 3 marks each.)

Question 26.
Elaborate the oxidizing character of KMnO₄ in:
(A) Alkaline solutions
(B) Neutral solutions [3]
Answer:
(A) In the presence of an alkali, purple coloured potassium permanganite reduces to green coloured potassium manganite. Potassium manganate further gets reduces to Mn02 in presence of reducing agent.
4 KMnO₄ + 2H₂O → 4MnO₂ + 3O₂ + 4KOH
It oxidises potassium iodide to potassium iodate.
It oxidises nitrotoluene to nitrobenzoic acid.

(B) In this, potassium permanganate behaves as moderate oxidizing agent as shown in the reaction:
MnO₂ + 2H₂O + 3e^– → MnO_2(s) + 4OH^–
It oxidises manganous sulphate to manganese dioxide.
It oxidises hydrogen sulphide to sulphur.

Question 27.
Write down the basic strength of following amines in increasing order. [3]
(A) o-nitro aniline, m-nitro aniline , p-nitro aniline, aniline
Answer:

(B) Methylamine, ethylamine, aniline , benzylamine
Answer:

(C) N, N-dimethyl aniline, aniline, N-methyl aniline
Answer:

Caution:
Students usually get confused between the acidic strength and basic strength of compounds. To answer this appropriately, student must understand the effect of electron-withdrawing and electron-releasing group towards the benzene and alkyl groups.

Question 28.
Account for any three of the following: [3]
(A) Transition metals and their compounds are known to be the good catalysts in many of the processes.
(B) Zn, Cd and Hg are not regarded as transition elements.
(C) Ce³⁺ can be easily oxidized to Ce⁴⁺
(D) Cr²⁺ is a stronger reducing agent than Fe²⁺
Answer:
(A) The capacity of transition metals and their compounds to change oxidation state or, in the case of the metals, to adsorb other substances on to their surface and activate them as a result, makes them catalysts.

(B) Due to their electrical makeup, elements like Zn, Cd, and Hg are not transition elements. These three parts’ electrical arrangement can be represented generally as (n – 1)d ¹⁰ ns². Both in their ground state and in their overall oxidation state, these elements’ orbitals are all filled. As a result, these components are not transitional elements.

(C) Ce⁴⁺ ions are formed when Ce³⁺ ions with the configuration 4f¹⁵d⁰6d⁰ lose one electron and change to the configuration 4f⁰5d⁰6s⁰.

(D) This is because E°(Cr³⁺/Cr²⁺): 3d⁵ 4s¹/ 3d⁵ is negative (-0.41V) whereas E°(Fe³⁺/Fe²⁺): (3d⁶/3d⁵) is positive (+0.77 V). Thus, Cr²⁺ is easily oxidised to Cr³ but Fe²⁺ cannot be easily oxidised to Fe³⁺.

Related Theory:
Oxidation = electron loss; Reduction = electron gain.
A reducing agent supplies electrons that go on to reduce another species. It is itself oxidised.

Question 29.
Write any three tests for glucose.
OR
Write a short note on:
(A) Mutarotation
(B) Similarity between D- Lactose and D-maltose
(C) Furanose structures of fructose [3]
Answer:
Test for glucose are as follow:
1. Bromine water test: when glucose is reacted with bromine water, the red color of bromine water disappears and solution turns colourless.

2. Silver mirror test: when glucose is warmed with ammonical silver nitrate, silver mirror appears on the inner walls of the test tube.

3. Fehlings solution test: when glucose is heated with Fehling’s solution, a red precipitate of Cu₂O is formed.

OR

(A) Due to a shift in the equilibrium between two anomers in a solution, mutarotation refers to a variation in the specific rotation of plane-polarized light. A hemiketal or hemiacetal group is required for any molecule to exhibit mutarotation. Sugar is where the mutarotation characteristic was originally discovered. Different sugars rotate in aqueous solutions in different ways.

(B) Both have α and □ anomers that can undergo mutarotation

(C) The fructofuranose structure is obtained by the internal ketal formation by combining the keto group at C₂ carbon and the -OH group at C₅ carbon.

Question 30.
Draw the structures of following complexes based on Werner’s theory and state the total no of ions participated. [3]
(A) CoCl₃.5NH₃
Answer:

Structure of COCl₃.5NH₃ complex
Number of Cl^– ions precipitated = 2
Total number of ions = 3

(B) CoCl₃.4NH₃
Answer:

Structure of COCl₃.4NH₃ complex
Number of Cl^– ions precipitated = 1
Total number of ions = 2

(C) CoCl₃.6NH₃
Answer:

Structure of COCl₃.6NH₃ complex
Number of Cl^– ions precipitated = 3
Total number of ions = 4

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.
Atoms and molecules interact throughout all chemical processes. A few grammes of each chemical composition contain many atoms or molecules, changing in number depending on their atomic or molecular masses. The mole idea was developed to efficiently handle such a big quantity. The mole idea is also the foundation of all electrochemical cell processes. For instance, 400 ml of an aqueous NaCl solution with a 5.0 molar concentration is electrolyzed. As a result, chlorine gas develops at one of the electrodes.

The mole idea may be used to determine how many items are produced:

(A) During reaction, what is the number of moles of Cl₂ gas evolved? [1]
(B) Name any two strong electrolyte in an aqueous solution. [1]
(C) (i) Write the unit of ∧_m.
(ii) Calculate ∧°_m for acetic acid:
Given that ∧°_m (HCl) = 426 S cm² mol^-1
∧°_m (NaCl) = 126 S cm² mol^-1
∧°_m (CH₃COONa) = 91 S cm² mol^-1
OR
A current of 3 amperes passing through AgNO₃ solution for 20 minutes deposited 4.0g of silver. Calculate electron chemical equivalent of silver (Atomic of silver = 108). [2]
Answer:
(A) Moles of NaCl = 5 × 0.4 = 2 moles.
2 moles of NaCl has 2 moles of Cl^–
2Cl → Cl₂ + 2e^–

The balanced chemical equation clearly depicts that 2 moles of Cl^– yields 1 mole of Cl₂ gas.
Therefore, 1 mole of Cl₂ gas has been evolved.

(B) AgNO₃, HCl

(C) (i) Sm² mol^-1

Explanation:
The unit of molar conductivity is Seimen m² mol^-1.

(ii) ∧°_m (CH₃COOH) = ∧°_m(H⁺) + ∧°m (CH₃COO^–) [Kohlraouch Law]
∧°_m (HCl) = ∧°_m(H⁺) + ∧°_m (Cl^–) …… (1)
∧°_m (NaCl) = ∧°_m(Na⁺) + ∧°_m (Cl^–) …… (2)
∧°_m (CH₃COONa) = ∧°m(CH₃COO^–) + ∧°_m (Na⁺) …. (3)
On subtracting 2 from 3 and on addition of 1 we get,
= 96 – 126 + 426
= 396
∧°_m (CH₃COOH) = 396 Scm² mol^-1

OR

Given that:
l = 3A
m = 4g
t = 20 minutes = 20 × 60 = 1200 seconds
Z = ?
We know that, m = Z × l × t
Also Z = \(\frac{m}{l}\) × t
Z = \(\frac{4}{3}\) × 1200
Z = \(\frac{1}{900}\) = 1.1 × 10^-3

Related Theory:
Chemical equivalent is the weight of the element corresponding to a unit weight of hydrogen, either as replacing it, or combining with it. It is often conveniently used in electrochemistry to avoid the necessity of dividing by the valency when atomic weights are used.

Question 32.
A hydroxy (-OH) group is directly joined to the aromatic ring in phenols. The term “phenolic group” often refers to a hydroxyl group that is joined to an aromatic ring. Phenols exhibit several reactions that are comparable to those of alcohols because they include the -OH group. However, because the -OH group is directly attached to the aromatic ring, it behaves quite differently from alcohols. Because of this, phenols act quite differently from alcohols in a variety of ways. The following categories can be used to group the chemical reactions of phenols.
(I) Responses involving the -OH group,
(II) Reactions using an aromatic ring, such as benzene.
(A) Which compounds are formed when phenol reacts with acid chlorides and acid anhydrides? [1]
(B) Which is the reason behind acidic character of phenols? [1]
(C) (i) What is the correct order of acidic strength of nitrophenols:
m-nitrophenol, o-nitrophenol, p-nitrophenol, phenol
(ii)

In the above reaction, which is the major product formed?

OR

Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl-ZnCl2 (b) HBr
(i) Butan-l-ol
(ii) 2-Methylbutan-2-ol [2]
Answer:
(A) Esters

Explanation:
Phenol reacts with acid chlorides and acid anhydrides and form esters. For example:

(B) The phenoxide ion also shows resonance. Both phenol and phenoxide ions are resonance stabilized but phenoxide ion acquires greater stability.

(C) (i) p-nitrophenol > o-nitrophenol > m-nitrophenol > phenol
Explanation:
The effect of electron-withdrawing groups such as -NO₂ group is more pronounced at o- and p- positions than m-positions.

(ii) Triphenyl phosphate
Explanation:

In the above reaction, the main product for the reaction is triphenyl phosphate.

OR

(i) With HCl-ZnCl₂

(ii) With HBr

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) When 0.0821g of hydroxyl benzaldehyde is dissolved in 20g of naphthalene. The freezing point of naphthalene is 80.1°C. and the depression in freezing point is found to be 0.232°C. then, find the molar depression constant and deduce the latent heat of fusion for naphthalene.
(B) Consider ΔT_b, is the elevation of boiling point and m is the molality of solution. Then, show that ΔT_b is directly proportional to m. (5)
Answer:
(A) Mass of solute (C₇H₆O₂), w = 0.0821 g
Mass of solvent naphthalene, W = 20 g
Depression of freezing point, ▢T_f = 0.232°C
Molar mass of solute (C₇H₆O₂), M’ = 84 + 6 +32 = 122

= 35.93 cat g.
Hence, the molal depression constant for
naphthalene is 6.895 K kg mol^-1 and its latent heat of fusion is 35.93 cat g^-1.

(B) An expression for the depression of the freezing point can be derived in the same way as used for the elevation of boiling point. The derivation is as follows.

If the curves AB, AC, BD and CE shown in Fig. above are assumed to be straight lines, triangles lines, triangles ABD and ACE may be regarded as similar triangles.
For similar triangles ABD and ACE, we have
\(\frac{\mathrm{BD}}{\mathrm{CE}}=\frac{\mathrm{AD}}{\mathrm{AE}}\)
or \(\frac{T_f-T_1}{T_f-T_2}\) = \(\frac{p^{\circ}-p_1}{p^{\circ}-p_2}\) …………. (1)
where, p° = Vapour pressure of liquid solvent at its freezing point T_f
P₁ = Vapour pressure of solution 1 at temperature T₁
p₂ = Vapour pressure of solution 2 at temperature T₂.
Eq. (i) can be written as
\(\frac{\Delta T_{f_1}}{\Delta T_{f_2}}=\frac{\Delta p_1}{\Delta p_2}\) …………….. (ii)
If follows from eq.(ii) that in general.
∆T_f ∝ ∆_p …………….. (iii)
Thus, depression of freezing point is directly proportional to the lowering of vapour pressure.
According to Raoult’s law, for a dilute solution,

Question 34.
Write down the following named reactions.
(A) Etard reaction.
(B) Gattermann- Koch reaction
OR
How will you distinguish between the
following?
(A) Benzaldehyde and acetophenone
(B) Pentan-2one and pentan-3-one
(C) Diethyl ketone and acetone
(D) Formaldehyde and acetaldehyde
(E) Acetaldehyde and benzaldehyde
Answer:
(A)

(B)

OR

(A) They can be distinguished by silver mirror test and iodoform test. Benzaldehyde gives silver mirror when heated with tollen’s reagent but acetophenone does not.

(B) They can be distinguished by iodoform test as pentan-2-one being a methyl ketone gives yellow ppt of iodoform while pentan- 3-one does not form yelLow ppt when heated with iodine and alkali.

(C) They are distinguished by sodium bisulphite test as acetone being a methyl ketone gives white ppt of bisulphite compound when treated with saturated solution of sodium bisulphite while diethyl ketone does not give this test.

(D) They are distinguished by iodoform test. Acetaldehyde gives yellow ppt of iodoform while formaldehyde does not give this test.

(E) They are distinguished by iodoform test as acetaldehyde gives this test while benzaldehyde does not.

Question 35.
(A) State yes or no. Can nickel displace hydrogen from hydrochloric acid. Explain your answer.
E°_Ni²⁺/Ni = -0.25 V
(B) State any four important features of electrochemical series.
(C) Draw the labelled diagram of Daniel cell and show:
(i) the reaction of half cells and the net cell reaction.
(ii) the direction of flow of current and direction of flow of electrons.
OR
(A) What are the criteria for product formation during electrolysis?
(B) State any three applications of Kohlrausch’s law.
(C) Name the metal that can be obtained by the electrolysis of an aqueous solution of its salts. [5]
Answer:
(A) Yes, nickel can displace hydrogen from hydrochloric acid. This is because, E° for the reaction is positive.
Ni_(s) + 2H⁺_(aq) → Ni²⁺_(aq) + H₂

(B) The features of electrochemical series are:

• The reactivity of metals decreases on moving down the group.
• Elements placed at the end of the series have the highest value of standard reduction potential.
• A metal can displace any other metal placed below it in the series from its salt solutions.
• The elements placed at the top of the electrochemical series have the minimum value of standard reduction potential.

(C)

Cr_(s) → Cr³⁺_(aq) + 3e^– Cr²⁺_(aq) + 2e^– → Cu(s)

OR

(A) The crieteria for product formation during electiolysis.

• The substance which possesses higher standard electrode potential is reduced at the cathode.
• The substance which has lower standard reduction potential is oxidized at the anode.

(B) Applications of Kohlrausch’s law are as follows:

• Determination of ∧_m^∞ for weak electrolytes.
• Determination of degree of dissociation of a weak electrolyte.
• Determination of the solubility of a sparingly soluble salt.

(C) Silver