Search Previous Year Questions

CONCEPT:

Coordination Chemistry and Precipitation Reaction

• The given reaction involves a coordination complex $[Cr(H_2O)_6]Cl_3$, which is a hexaaquachromium(III) chloride complex.
• When this complex reacts with an excess of AgNO₃ (silver nitrate), the chloride ions present outside the coordination sphere precipitate as AgCl (silver chloride).
• The number of moles of AgCl precipitated depends on the number of chloride ions outside the coordination sphere of the complex.

EXPLANATION:

• The reaction can be written as:

\( [Cr(H_2O)_6]Cl_3 + 3AgNO_3 \xrightarrow{excess} 3AgCl(s) + Cr(H_2O)_6_3 \)

• In the given coordination complex $[Cr(H_2O)_6]Cl_3$, all six water molecules are within the coordination sphere of the chromium ion, while the three chloride ions are outside the coordination sphere.
• When reacted with excess AgNO₃, each chloride ion forms a precipitate of AgCl:

$Cl^- + Ag^+ \rightarrow AgCl(s)$

• Thus, one mole of $[Cr(H_2O)_6]Cl_3$ reacts to produce 3 moles of AgCl precipitate.
• The ratio of the moles of the complex reacted to the moles of AgCl precipitated is given as:

$\frac{\text{no. of moles of complex reacted}}{\text{no. of moles of AgCl precipitated}} \times 100$

= $\frac{1}{3} \times 100$

= 33.3%

Therefore, the percentage of the moles of the complex reacted with respect to the moles of AgCl precipitated is 33.3%.

CONCEPT:

Mathematical Addition

• The addition of numbers is a fundamental arithmetic operation.
• It involves combining two or more numbers to find their total.
• The general form for addition is:

a + b = c, where c is the sum of a and b.

EXPLANATION:

Therefore, the sum of 1 + 4 + 4 + 2 + 3 is 14.

CONCEPT:

Combustion of Alkanes

• Alkanes are hydrocarbons with the general formula $C_nH_{2n+2}$.
• When alkanes undergo combustion in the presence of oxygen ($O_2$), they produce carbon dioxide ($CO_2$) and water ($H_2O$) as products.
• The balanced chemical equation for the combustion of an alkane is:

$C_nH_{2n+2} + \left( \frac{3n+1}{2} \right) O_2 \rightarrow nCO_2 + (n+1) H_2O$

EXPLANATION:

• In the given question, we are tasked with identifying the alkane based on the combustion equation.
• The given combustion equation includes the term $\left( \frac{3n+1}{2} \right)$ for the oxygen ($O_2$) coefficient.
• We are also provided with the condition $\frac{3n+1}{2} = 8$.
• Solving this equation:
• $\frac{3n+1}{2} = 8$
• $3n+1 = 16$ (multiplying both sides by 2)
• $3n = 15$ (subtracting 1 from both sides)
• $n = 5$ (dividing both sides by 3)
• Since $n = 5$, the alkane is $C_5H_{12}$ (pentane).
• Substituting $n = 5$ into the combustion equation:

$C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O$

Therefore, the alkane is pentane ($C_5H_{12}$), and the number of oxygen molecules required for complete combustion is 8.

CONCEPT:

Arrhenius Equation and Activation Energy

• The Arrhenius equation relates the rate constants of a reaction at two different temperatures with the activation energy of the reaction.
• The equation is given as:

ℓn(K2K1)=EaR[T2−T1T1T2]ℓn(K2K1)=EaR[T2−T1T1T2]

• Here:
• K1K1

and K2

are the rate constants at temperatures T1

and T2

respectively.

• Ea

is the activation energy of the reaction.

• R

is the universal gas constant (8.3 J mol−1 K−1

).

• T1

and T2

are the absolute temperatures in Kelvin.

EXPLANATION:

• In the given question:
• K1=1.5×103

, K2=4.5×103

• Ea=60×103 J mol−1

, T1=300 K

• Substitute these values into the Arrhenius equation:

ℓn(4.5×1031.5×103)=60×1038.3[T2−300300.T2]

• Simplify the natural logarithm:
• ℓn(4.5×1031.5×103)=ℓn 3
• ℓn 3=1.0986
• Rearrange the equation:

1.0986=60×1038.3[T2−300300.T2]

• Calculate:
• 60×1038.3=7228.92
• 1.0986=7228.92[T2−300300.T2]
• T2−300300.T2=1.09867228.92
• T2−300300.T2=0.000152
• T2−300=0.000152⋅300.T2
• T2−300=0.0456⋅T2
• T2−0.0456⋅T2=300
• 0.9544⋅T2=300
• T2=3000.9544
• T2=314.4 K
• Convert T2

to Celsius:

• t=T2−273
• t=314.4−273
• t=41.4∘C

Therefore, the temperature T2

is 314.4 K, which corresponds to 41.4∘C

.

CONCEPT:

Equilibrium and Gibbs Free Energy

• At equilibrium under standard pressure conditions, the Gibbs free energy change ($\Delta G^\circ$) is zero.
• The relationship between Gibbs free energy and equilibrium temperature can be expressed using thermodynamic principles.

EXPLANATION:

• The given equation is:

105 – 35 log T = 0

• We solve this equation to find the value of temperature ($T$):
• Rearranging the equation:

35 log T = 105

• Dividing both sides by 35:

log T = 105 / 35

log T = 3

• Converting from logarithmic to exponential form:

$T = 10^{3}$

$T = 1000 , \text{K}$

• Converting temperature from Kelvin to Celsius:

$T (\text{in } ^\circ \text{C}) = T (\text{in K}) - 273$

$T (\text{in } ^\circ \text{C}) = 1000 - 273$

$T (\text{in } ^\circ \text{C}) = 727^\circ \text{C}$

Therefore, the temperature at equilibrium is $T = 1000 , \text{K}$ or $T = 727^\circ \text{C}$.

Given that:

$\alpha, \alpha + 2, \alpha \in \mathbf{Z}$, are the roots of the quadratic equation:

$x(x + 2) + (x + 1)(x + 3) + (x + 2) (x + 4) + \ldots + (x + n - 1) (x + n + 1) = 4n$, where $n \in \mathbf{N}$.

Calculation:

$nx^2 + x(2 + 4 + 6 + \ldots + 2n) + (1 \cdot 3 + \ldots + (n - 1)(n + 1)) = 4n$

⇒ $nx^2 + n(n + 1)x + \frac{n(n - 1)(2n + 5)}{6} = 4n$

⇒ $x^2 + (n + 1)x + \frac{(n - 1)(2n + 5)}{6} = 4$

Discriminant must be a perfect square:

$D = (n + 1)^2 - 4 \times \frac{(n - 1)(2n + 5)}{6}$

⇒ $D = \frac{122 - 2n^2}{6} = 20 - \left( \frac{n^2 - 1}{3} \right)$

For a perfect square:

$\frac{n^2 - 1}{3} = 16$

⇒ $n = 7$

Substitute $n = 7$ into the equation:

$x^2 + 8x + \frac{8 \times 15}{6} - 5 = 0$

⇒ $x^2 + 8x + 15 = 0$

Roots: $x = -3, -5$

$\alpha = -5, \alpha + 2 = -3$

$\alpha + n = 7 - 5 = 2$

∴ The correct answer is option (3).

Given that:

$50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i$

Calculation:

$50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i$

⇒ $\frac{1}{1+3i}=\frac{1-3i}{10}$ and $\frac{1}{1-2i}=\frac{1+2i}{5}$

⇒ $10x-30xi-10y-20yi=31+17i$

⇒ Real part: $10(x-y)=31$, Imaginary part: $-30x-20y=17$

⇒ $x=0.9,;y=-2.2$

⇒ $10(x-3y)=75$

∴ The correct answer is option (4).

Given that:

$x + 2y + z = 5$

$2x + y + \alpha z = 5$

$8x + 4y + \beta z = 18$

Calculation:

$\Delta = \begin{vmatrix} 1 & 2 & 1 \ 2 & 1 & \alpha \ 8 & 4 & \beta \end{vmatrix} = 0$

$\Delta = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(0) = 0$

$\beta - 4\alpha - 4\beta + 16\alpha = 0$

$-3\beta + 12\alpha = 0$

$4\alpha = \beta$

$\frac{\beta}{\alpha} = 4$

∴ The correct answer is option (2).

Given that:

$A = \begin{bmatrix} 1 & 2 \ 1 & \alpha \end{bmatrix}, ; A^2 - 4A + I = O$

$B = \begin{bmatrix} 3 & 3 \ \beta & 2 \end{bmatrix}, ; B^2 - 5B - 6I = O$

Calculation:

$\text{Trace of } A:$; $\alpha + 1 = 4 \Rightarrow \alpha = 3$

$\text{Determinant of } B:$; $\text{det(B)} = 6 - 3\beta = -6 \Rightarrow 3\beta = 12 \Rightarrow \beta = 4$

B - A = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix}, ; B + A = \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix}

(B - A)(B + A) = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 13 & 15 \ 7 & 10 \end{bmatrix}

\text{adj(A + B)} = \begin{bmatrix} 5 & -5 \ -5 & 4 \end{bmatrix}, ; $\text{det(adj(A + B))} =$ (5)(4) - (-5)(-5) = 20 - 25 = -5

∴ The correct answer is option (3).

Given that:

$\text{AP}_1 \text{ (Set A)} = { 1, 6, 11, 16, \dots, 101 \text{ terms} }$

$\text{AP}_2 \text{ (Set B)} = { 9, 16, \dots, 71 \text{ terms} }$

$D = \text{L.C.M. } { d_1, d_2 } = 35$

$1^{\text{st}} \text{ Common term} = 16$

Calculation:

$16 + (n - 1) \cdot 35 \leq 499$

⇒ $n \leq 14.8$

⇒ $n = 14$

$A \cap B = { 16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471 }$

Divisible by 3:

$= { 51, 156, 261, 366, 471 }$

∴ The correct answer is option (2).