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CONCEPT:

• The volume of CO₂ gas can be determined using the moles of CO₂ gas in the sealed container.
• The number of moles of CO₂ is derived from the neutralization of the unreacted excess Ca(OH)₂ with HCl.

EXPLANATION:

• Let moles of CO₂ be n .
• Since the entire amount of CO₂ reacts with exactly half the initial amount of Ca(OH)₂, the initial moles of Ca(OH)₂ must be 2n .
• The excess Ca(OH)₂ remaining would then be n moles.
• The gram equivalents of Ca(OH)₂ remaining must equal the gram equivalents of HCl used to neutralize it.
• n x 2 = 0.1 x 40/1000 $\times 1$

2n = 0.004

n = 0.002

• Setting up the equation:
• This simplifies to:

The moles of CO₂ gas is 0.002 moles.

Using the molar volume of an ideal gas (22,400 cm³ at STP):

$0.002 \times 22,400 = 44.8 \text{ cm}^3$

• Volume of CO₂:
• Nearest integer: 45 cm³

Therefore, the volume of CO₂ is 45 cm³

CONCEPT:

Carius Method for Estimation of Halogens

• In this method, an organic compound containing halogens is treated with silver nitrate to form silver halide precipitate.
• The mass of silver halide is used to determine the amount of halogen present in the organic compound.

EXPLANATION:

• Mass of organic compound = 180 mg = 0.18 g
• Mass of AgCl produced = 143.5 mg = 0.1435 g
• Molar mass of Ag = 108 g/mol
• Molar mass of Cl = 35.5 g/mol
• Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol

To find the moles of AgCl produced:

• n(AgCl) = mass / molar mass
• n(AgCl) = 0.1435 g / 143.5 g/mol
• n(AgCl) = 0.001 mol

Since 1 mol of AgCl contains 1 mole of Cl:

• n(Cl) = n(AgCl) = 0.001 mol

Mass of Cl in the compound:

• mass(Cl) = n(Cl) × molar mass of Cl
• mass(Cl) = 0.001 mol × 35.5 g/mol
• mass(Cl) = 0.0355 g = 35.5 mg

Percentage of Cl in the compound:

• %Cl = (mass(Cl) / mass of organic compound) × 100
• %Cl = (35.5 mg / 180 mg) × 100
• %Cl ≈ 19.72%

Rounding to the nearest specified range:

• The percentage composition of chlorine in the compound is 20%

CONCEPT:

Linear Geometry

• Linear geometry occurs when the bond angle between atoms is 180°.
• Linear molecules or ions are typically formed when:
• The central atom has no lone pairs and is surrounded by two bonding domains (sp hybridization).
• The central atom has lone pairs but achieves a linear shape due to symmetric electron distribution (e.g., sp³d hybridization).

EXPLANATION:

• BeCl₂: Linear geometry due to sp hybridization (Cl–Be–Cl).

• CO₂: Linear geometry due to sp hybridization (O=C=O).

• N₃⁻: Linear geometry due to sp hybridization (N=N=N).

• XeF₂: Linear geometry due to sp³d hybridization (F–Xe–F, with three lone pairs in equatorial positions).

• NO₂⁺: Linear geometry due to sp hybridization (O=N=O).

• I₃⁻: Linear geometry due to sp³d hybridization (I–I–I, with three lone pairs in equatorial positions).

• Other species such as SO₂, NO₂, and O₃ have bent geometries.

Therefore, the number of species with linear geometry is 6: BeCl₂, CO₂, N₃⁻, XeF₂, NO₂⁺, and I₃⁻.

Ans. (69)  

Sol.

$t_{1 / 2}=\frac{0.693}{K}$​

K = Ae^–Ea/RT

= $10^{20} \times \mathrm{e}^{-\frac{191.48 \times 10^{3}}{8.314 \times 1000}}$

= 10²⁰ × e^–23.031 = 10²⁰ × – e^{ln 10 × 10}

= $\frac{10^{20}}{10^{10}}=10^{10} \mathrm{sec} .$

$\mathrm{t}_{1 / 2}=\frac{0.693}{10^{10}}=6.93 \times 10^{-11}$

= 69.3 × 10^–12 sec

CONCEPT:

First Order Kinetics

• The time required for half of the reactant to convert to product (half-life, t_1/2) can be determined using the rate constant (k).
• The Arrhenius equation relates the rate constant to the temperature and activation energy: $k = A e^{-\frac{E_a}{RT}}$

EXPLANATION:

• Given:
• Activation energy, E_a = 191.48 kJ/mol = 191,480 J/mol
• Frequency factor, A = 10²⁰ s⁻¹
• Temperature, T = 1000 K
• Universal gas constant, R = 8.314 J K⁻¹ mol⁻¹

We first solve for the rate constant (k) using the Arrhenius equation:

• $k = A e^{-\frac{E_a}{RT}} \ k = 10^{20} e^{-\frac{191,480}{8.314 \cdot 1000}} \ k = 10^{20} e^{-\frac{191.48}{8.314}} \ k = 10^{20} e^{-23.031} \ e^{-23.031} \approx 10^{-10} (since ln(10) \approx 2.303 , result closely matches 10^{-10} ) \ k = 10^{20} \cdot 10^{-10} \ k = 10^{10} \text{ s}^{-1}$

Using the rate constant to find the half-life (t_1/2) for first-order kinetics:

• $t_{1/2} = \frac{0.693}{k} \ t_{1/2} = \frac{0.693}{10^{10} \text{ s}^{-1}} \ t_{1/2} = 6.93 \times 10^{-11} \text{ s}$
• Converting to picoseconds (1 s = 10¹² ps): $t_{1/2} = 6.93 \times 10^{-11} \text{ s} \times 10^{12} \frac{\text{ps}}{\text{s}}$
• $t_{1/2} = 69.3 \text{ ps}$

Rounding to the nearest specified range:

• The time required for 50% of molecules of A to become B is 69 picoseconds

CONCEPT:

Reaction Mechanism Involving Reduction, Diazotization, and Sandmeyer Reaction

• The given sequence of reactions involves multiple steps:
• Reduction: The nitro group (-NO₂) is reduced to an amino group (-NH₂) using Sn and HCl.
• Diazotization: The amino group is converted into a diazonium salt (-N₂⁺) using NaNO₂ and HCl at 0-5°C.
• Sandmeyer Reaction: The diazonium salt reacts with Cu₂Cl₂, replacing the diazonium group with a chlorine atom.
• Wurtz Coupling: Chlorobenzene undergoes a reaction with sodium in ether to form biphenyl.

EXPLANATION:

• Step 1: Reduction
• Benzene with a nitro group (C₆H₅NO₂) reacts with Sn and HCl to form aniline (C₆H₅NH₂).
• Step 2: Diazotization
• Aniline reacts with NaNO₂ and HCl at 0-5°C to form benzene diazonium chloride (C₆H₅N₂Cl).
• Step 3: Sandmeyer Reaction
• Benzene diazonium chloride reacts with Cu₂Cl₂ to form chlorobenzene (C₆H₅Cl).
• Step 4: Wurtz Coupling
• Chlorobenzene reacts with sodium in ether to form biphenyl (C₆H₅–C₆H₅).

Molar mass of biphenyl (C₁₂H₁₀)

• Carbon (C): 12 atoms × 12 g/mol = 144 g/mol
• Hydrogen (H): 10 atoms × 1 g/mol = 10 g/mol
• Total molar mass = 144 + 10 = 154 g/mol

Therefore, the molar mass of the product (A) is 154 g/mol.

Concept Used:

The number of equivalence relations on a set is equivalent to the number of ways the set can be partitioned. For a set containing n elements, this number is represented by the Bell number Bn.

Calculation:

⇒ Partitions of A = {1, 2, 3} are:

⇒ 1. {{1}, {2}, {3}}

⇒ 2. {{1, 2}, {3}}

⇒ 3. {{1, 3}, {2}}

⇒ 4. {{2, 3}, {1}}

⇒ 5. {{1, 2, 3}}

⇒ Number of partitions = 5

⇒ Bell number B3 = 5

⇒ Number of equivalence relations = 5

The empty relation is not included

∴ The maximum number of equivalence relations on A = {1, 2, 3} is 5

Hence option 3 is correct

Calculation 

f(x + y) = f(x).f(y) 

⇒ f(x) = e^λx f′(0) = 4a 

⇒ f′(x) = λe^λx ⇒ λ = 4a

So, f(x) = e^4ax

f′′(x) – 3af′(x) – f (x) = 0 

⇒ λ² – 3aλ – 1 = 0 

⇒ 16a² – 12a² – 1 = 0 ⇒ 4a² = 1 ⇒ $\mathrm{a}=\frac{1}{2}$

F(x) = e^2x

Area = $\int_{0}^{2} \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{e}^{2}-1$

Hence option 1 is correct

Calculation 

Let ‘G’ be the centroid of Δ formed by (1, 3) (3, 1) & (2, 4)

$\mathrm{G} \cong\left(2, \frac{8}{3}\right)$

Image of G w.r.t. x + 2y – 2 = 0

$\frac{α-2}{1}=\frac{β-\frac{8}{3}}{2}=-2 \frac{\left(2+\frac{16}{3}-2\right)}{1+4}$

= $\frac{-2}{5}\left(\frac{16}{3}\right)$

⇒ $α=\frac{-32}{15}+2=\frac{-2}{15}, β=\frac{-32 \times 2}{15}+\frac{8}{3}=\frac{-24}{15}$

15(α – β) = – 2 + 24 = 22

Hence option 4 is correct

Calculation

Given:

$Z_{1}=e^{-i\pi/4}, Z_{2}=1, Z_{3}=e^{i\pi/4}$

\( |z_{1}\overline{z}{2}+z{2}\overline{z}{3}+z{3}\overline{z}_{1}|^{2} = |e^{-i\frac{\pi}{4}}\times1+1\times e^{-i\frac{\pi}{4}}+e^{i\frac{\pi}{4}}\times e^{i\frac{\pi}{4}}|^{2} \)

$= |e^{-i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}}+e^{i\frac{\pi}{4}}|^{2}$

$= |2e^{-i\frac{\pi}{4}}+1|^{2}$

$= |\sqrt{2}-\sqrt{2}i+i|^{2}$

$= (\sqrt{2})^{2}+(1-\sqrt{2})^{2} = 2+1+2-2\sqrt{2} = 5-2\sqrt{2}$

$\alpha=5, \beta=-2$

$\Rightarrow \alpha^{2}+\beta^{2}=29$

Hence option (4) 29 is correct

Calculation 

16(sec^–1 x)² + (cosec^–1 x)²

\operatorname{Sec}^{-1} \mathrm{x}=\mathrm{a} \in[0, π]-\left{\frac{π}{2}\right}

$\operatorname{cosec}^{-1} \mathrm{x}=\frac{π}{2}-\mathrm{a}$

= $16\left[\mathrm{a}^{2}+\left(\frac{π}{2}-\mathrm{a}\right)^{2}\right]=16\left[2 \mathrm{a}^{2}-π \mathrm{a}+\frac{π^{2}}{4}\right]$

$\max ]_{a=π}=16\left[2 π^{2}-π^{2}+π \frac{2}{4}\right]=20 π^{2}$

$\min ]_{a=\frac{π}{4}}=16\left[\frac{2 \times π^{2}}{16}-\frac{π^{2}}{4}+\frac{π^{2}}{4}\right]=2 π^{2}$

Sum = 22π² 

Hence option 4 is correct