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CONCEPT:

Electrolysis of Water and Expansion Work

• The electrolysis of water breaks down 2 moles of water (H₂O) into 2 moles of hydrogen (H₂) and 1 mole of oxygen (O₂), producing 3 moles of gas per 2 moles of water electrolysis.
• The expansion work done during this process can be calculated using the formula:

W = -P ΔV

where P is the pressure and ΔV is the change in volume.

• For an ideal gas, the change in volume (ΔV) during the expansion can be calculated using the ideal gas law:

ΔV = nRT / P

where n is the number of moles of gas produced, R is the universal gas constant, T is the temperature in Kelvin, and P is the pressure.

• For the electrolysis of water, the number of moles of gas produced is based on the molar volume at the given conditions, and the work done during the process can be calculated.

EXPLANATION:

• First, calculate the moles of water:

moles of water = 144 g / 18 g/mol = 8 moles

• From the electrolysis reaction, 2 moles of water produce 3 moles of gas (H₂ and O₂). Therefore, 8 moles of water produce 12 moles of gas.
• The expansion work is calculated as:

W = -nRT

where n = 12 moles of gas, R = 8.3 J/mol·K, and T = 300 K:

W = -(12 × 8.3 × 300) = -29,880 J = -29.88 kJ

Therefore, the expansion work done is -29.88 kJ.

CONCEPT:

Monomer Involved in Nylon 6,6 Synthesis and Nitrogen Gas Evolution

• The monomer X involved in the synthesis of Nylon 6,6 undergoes a positive carbylamine test, which indicates the presence of a primary amine group (-NH₂).
• The molecular formula of the monomer X is assumed to be X = NH₂(CH₂)₆NH₂, indicating it is a diamine.
• When 10 moles of X are analyzed using Dumas' method, the amount of nitrogen gas evolved can be calculated. In Dumas' method, nitrogen gas evolves due to the decomposition of the amine group.
• The relationship between moles of the monomer and nitrogen gas produced is 1:1, as each mole of X releases 1 mole of nitrogen gas (N₂).

EXPLANATION:

• Given that 10 moles of X (NH₂(CH₂)₆NH₂) are analyzed, we know that 1 mole of X releases 1 mole of nitrogen gas (N2).
• The molar mass of nitrogen (N) is 14 g/mol, so 1 mole of nitrogen gas weighs 28 g (since N2 is diatomic).
• Therefore, 10 moles of X will produce 10 moles of nitrogen gas, which will have a mass of:

10 moles × 28 g/mol = 280 grams

Therefore, the amount of nitrogen gas evolved is 280 grams.

CONCEPT:

Reaction Sequence and Product Yield Calculation

• The reaction involves a series of steps, starting with the compound X, and in each step, the yield of the major product is given.
• We are asked to determine the amount (in grams) of product S produced, based on the yields and stoichiometry of the reaction steps.
• The molar masses of the atoms are provided:
• H = 1
• C = 12
• O = 16
• Br = 80

EXPLANATION:

• The reaction sequence involves the following steps:
• Step 1: X undergoes reaction with NaOH to form P with a yield of 50%. 16 moles of X lead to 8 moles of P.
• Step 2: P reacts with NaOH, CaO to form Q with a yield of 50%. 8 moles of P lead to 4 moles of Q.
• Step 3: Q reacts with PbBr₂ to form R with a yield of 50%. 4 moles of Q lead to 2 moles of R.
• Step 4: R reacts with NH₃ to form S with a yield of 50%. 2 moles of R lead to 1 mole of S.
• The mass of S produced:
• The molar mass of S can be calculated by adding the atomic masses of the constituent elements in S.
• S is formed from the compound R, and its structure is shown in the image. The molar mass of S is calculated as follows:

Molar mass of S = (26 × 12) + (16 × 1) + (1 × 22) = 350 g/mol

• Since 1 mole of S is produced from 1 mole of R, and 1 mole of R is produced from 2 moles of Q, and 4 moles of P are used to make 2 moles of Q, the total amount of S produced is:

Amount of S = 0.5 moles × 350 g/mol = 175 g

Therefore, the amount of product S produced is 175 grams.

CONCEPT:

Precipitation of Metal Ions using Group Reagents

• Group reagents are chemicals that are used to selectively precipitate metal ions from their solutions based on their solubility and reactivity.
• The process typically involves adding a reagent that forms an insoluble compound with a specific metal ion. The ions that precipitate are usually those with the least solubility in the given conditions.
• For example, adding H₂S in the presence of NH₄OH can precipitate metal ions like Mn²⁺ as MnS.
• Similarly, other reagents such as (NH₄)₂CO₃, NH₄OH, and HCl can be used to precipitate different metal ions such as Ba²⁺, Al³⁺, Cu²⁺, etc.

EXPLANATION:

• P: Passing H₂S in the presence of NH₄OH
• This reagent combination is used to precipitate Mn²⁺ as MnS.
• Thus, P → 3 (Mn²⁺)
• Q: (NH₄)₂CO₃ in the presence of NH₄OH
• This reagent combination precipitates Ba²⁺ as BaCO₃.
• Thus, Q → 4 (Ba²⁺)
• R: NH₄OH in the presence of NH₄Cl
• This reagent combination is used to precipitate Al³⁺ as Al(OH)₃.
• Thus, R → 2 (Al³⁺)
• S: Passing H₂S in the presence of dilute HCl
• This reagent combination is used to precipitate Cu²⁺ as CuS.
• Thus, S → 1 (Cu²⁺)

Therefore, the correct match is P → 3 ; Q → 4 ; R → 2 ; S → 1

CONCEPT:

Organic Reactions and Their Products

• Each organic reaction has a specific reagent or set of reagents that lead to the formation of particular products. The reagents used in these reactions are crucial to determining the product formed in each case.
• For example, in the Stephen reaction, the conversion of toluene (methylbenzene) to benzaldehyde involves the use of CrO₃ in CS₂ or H₂O⁺. In the Sandmeyer reaction, a diazonium salt undergoes halogenation with CuX.
• In the Cannizzaro reaction, a non-enolizable aldehyde undergoes disproportionation in the presence of strong bases, while the Hoffmann bromamide degradation reaction involves the formation of amines from amides.

EXPLANATION:

• P: Stephen reaction – This reaction involves the oxidation of toluene to benzaldehyde using CrO₃ in CS₂ or H₂O⁺. The correct corresponding entry is (1) Toluene with the reagents (i) CrO₃, Cl₂/CS₂ or (ii) H₂O⁺.

• Q: Sandmeyer reaction – In this reaction, the diazonium salt of an aromatic amine is treated with CuX to form the corresponding halide. The corresponding entry is (3) Nitrobenezene with reagents (i) PC₁ or (ii) NH₃, (iii) P₄O₁₀, Δ.

• R: Hoffmann bromamide degradation reaction – This reaction involves the degradation of amides to primary amines by the use of bromine in the presence of NaOH. The correct corresponding entry is (2) Benzoic acid with reagents (i) Fe, HCl, or (ii) HCl, NaNO₃, (273–278 K), H₂O.

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When benzamide is heated with phosphorus pentoxide (P4O10), it undergoes a dehydration reaction to produce benzonitrile

• S: Cannizzaro reaction – This reaction involves the disproportionation of non-enolizable aldehydes in the presence of strong bases to form a mixture of alcohol and acid. The correct corresponding entry is (4) Toluene with reagents (i) Cl₂, hv, H₂O or (ii) Tollen’s reagent.

Therefore, the correct match is Option 2: P → 2 ; Q → 3 ; R → 4 ; S → 1.

CONCEPT:

Reactions and Tests for Organic Compounds

• Certain organic compounds undergo characteristic reactions with specific reagents, giving rise to distinct observations such as color changes or the formation of precipitates.
• For example, phenyl diazonium salts react with compounds containing amino or hydroxyl groups to produce dyes of specific colors. Similarly, compounds containing hydrazine groups react with glucose to give corresponding hydrazones.
• Each compound in List-I has a unique reaction with a corresponding reagent, as indicated in List-II.

EXPLANATION:

• R: Reaction with phenyl diazonium salt gives yellow dye.

• This compound is an amino acid derivative and gives the ninhydrin test. The yellow dye formation is characteristic of the reaction of amino acid derivatives with phenyl diazonium salts.

• Thus, P → 1 (Reaction with phenyl diazonium salt gives yellow dye).

• P: Reaction with ninhydrin gives purple color and it also reacts with FeCl₃ to give violet color.

• This compound is an amino acid derivative that reacts with ninhydrin to give a purple color. It also reacts with FeCl₃ to give a violet color, indicating the presence of a phenol group.

• Thus, Q → 5 (Reaction with ninhydrin gives purple color and reacts with FeCl₃ to give violet color).

• S: Reaction with glucose will give corresponding hydrazone.

• This compound contains a hydrazine group and reacts with glucose to form a hydrazone, which is a common reaction for hydrazines.

• Thus, R → 3 (Reaction with glucose gives corresponding hydrazone).

Therefore, the correct match is Option B: P → 1 ; Q → 5 ; R → 3 ; S → 4.

Concept:

• Polynomial Functions: These are algebraic expressions involving powers of the variable x. The highest power defines the degree.
• Function Transformation: When evaluating f(x + 1) or g(x + 2), we shift the graph of the function horizontally.
• Equal Functions: If f(x) ≠ g(x) ∀ x ∈ ℝ, then their difference is a non-zero function ∀ x.
• Coefficient Comparison: Equating coefficients of like powers of x in polynomials helps find unknown constants.

Calculation:

Let f(x) = a₁ + 10x + a₂x² + a₃x³ + x⁴

Let g(x) = b₁ + 3x + b₂x² + b₃x³ + x⁴

Define h(x) = f(x + 1) − g(x + 2)

Given: f(x) ≠ g(x) ∀ x ∈ ℝ

⇒ f(x) − g(x) ≠ 0 ∀ x

⇒ (a₃ − b₃)x³ + (a₂ − b₂)x² + 3x + a₁ − b₁ = 0

This equation has no real solution ⇔ Identically zero only if:

⇒ a₃ = b₃, and quadratic (a₂ − b₂)x² + 7x + a₁ − b₁ has no real roots.

Now, h(x) = (x + 1)⁴ + a₃(x + 1)³ + a₂(x + 1)² + 10(x + 1) + a₁ − (x + 2)⁴ − b₃(x + 2)³ − b₂(x + 2)² − 3(x + 2) − b₁

Coefficient of x³ in h(x):

= Coeff. of x³ from (x + 1)⁴ + a₃(x + 1)³ − (x + 2)⁴ − b₃(x + 2)³

= 4 + a₃ × 1 − 8 − b₃ × 1

= 4 + a₃ − 8 − b₃ = −4 (Given: a₃ = b₃)

∴ The coefficient of x³ in h(x) is −4.

Concept:

• Probability of Union of Events: The probability of the union of events A, B, C is P(A ∪ B ∪ C).
• Intersection of Events: The intersection represents common outcomes. Denoted as P(A ∩ B).
• Venn Diagram: A useful visual to represent unions, intersections, and complements of sets/events.
• Total Probability: The sum of all individual and overlapping probabilities in the sample space equals 1.
• Fraction Algebra: Used for solving unknowns by setting up equations with fractions.

Calculation:

P(U) = P(S₁ ∪ S₂ ∪ S₃) = 1/5

P(V) = P(S₁ / (S₂ ∪ S₃)) = 1/10

Let P(S₁ ∩ (S₂ ∪ S₃)) = x

P(V) = x / (0.5 + x) = 1/10

⇒ 10x = 0.5 + x

⇒ 9x = 0.5

⇒ x = 5/90 = 1/18

P(T) = 1/2 − 1/12 − 1/18

P(T) = 13/36

∴ Final Answer: P(T) = 13/36

Concept:

• Continuity at a Point: A function f(x) is said to be continuous at x = a if lim_x→a f(x) = f(a).
• Differentiability at a Point: A function is differentiable at x = a if both left-hand and right-hand derivatives exist and are equal.
• Left-hand Limit (LHL): lim_h→0⁺ f(a + h) must be finite and match the derivative from the right side.
• Trigonometric Limits: Use lim_x→0 sin(1/x) does not exist but x·sin(1/x) → 0 as x → 0.

Calculation:

f(x) = \(\left{\begin{array}{cc} 2-2 x^{2}-x^{2} \sin \frac{1}{x} & \text { if } x \neq 0 \ 2 & \text { if } x=0 \end{array}\right.\) 

f(x) is continuous at x = 0 and differentiable at x = 0

f′(x) = −4x − 2x·sin(1/x) + cos(1/x)

⇒ f′(x) = −2x[2 + sin(1/x)] + cos(1/x)

Now,

lim_h→0⁺ f(h) = lim_h→0⁺ [2 − 2h² − h²·sin(1/h)]

= lim_h→0⁺ [2 − h²(2 + sin(1/h))]

⇒ lim_h→0⁺ f(h) = 2

∴ f(x) is decreasing from x = 0 to some positive x = δ.

Hence Option 2 is the correct answer

Concept:

• Orthogonal Matrix: A square matrix Q is orthogonal if Q × Q^T = I, where I is the identity matrix and Q^T is the transpose of Q.
• Matrix Multiplication Property: If matrices P and Q commute (i.e., PQ = QP), then they satisfy certain structural constraints.
• Zero Pattern Multiplication: When multiplying matrices with zeroes in known positions, use simplification by directly eliminating multiplication with zero elements.
• Identity Matrix (I): A diagonal matrix with 1’s on the diagonal and 0 elsewhere. Acts as a multiplicative identity for matrices.

Calculation:

P = \(\left(\begin{array}{lll} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 3 \end{array}\right)\) 

Let Q = \(\left(\begin{array}{lll} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{array}\right)\)  

Given: PQ = QP

From multiplication, equating PQ and QP leads to constraints:

a₃ = 0, b₃ = 0, c₁ = 0, c₂ = 0

Thus Q becomes: \(\left(\begin{array}{lll} a_1 & a_2 & 0 \ b_1 & b_2 & 0 \ 0 & 0 & c_3 \end{array}\right)\) 

Q is orthogonal ⇒ Q × Q^T = I

a₁² + a₂² = 1

b₁² + b₂² = 1

a₁b₁ + a₂b₂ = 0

c₃² = 1

⇒ c₃ = ±1

Now two cases:

Case 1: a² = 1, b² = 0 ⇒ d² = 0 ⇒ 1 solution

Case 2: a² = 0, b² = 1 ⇒ d² = 0 ⇒ 1 solution

Case 3: a² = 1/2, b² = 1/2, and a₁b₁ + a₂b₂ = 0 ⇒ 4 solutions

So total number of Q matrices = 2 × (4 + 4) = 16

∴ Total number of orthogonal matrices Q is 16.