Class 11th

Class 11th Chemistry Some Basic Concepts of Chemistry

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0.2g of an organic compound was subjected to estimation of nitrogen by Dumas method in which volume of N₂ evolved (at STP) was found to be 22.400 mL. The percentage of nitrogen in the compound is __________. [nearest integer]

(Given : Molar mass of N₂ is 28g mol^-1, Molar volume of N₂ at STP : 22.4L)

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Number of moles of gas = $\frac{22.4}{22400} = \frac{1}{1000} m o l e s$

Mass of nitrogen = $(\frac{1}{1000} * 28) g m$

% of N in organic compound = $\frac{(\frac{28}{1000})}{0.2} * 100 % = 14 %$

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The set of all values of k for which ${(t a n^{- 1} x)}^{3} + {(c o t^{- 1} x)}^{3} = k π^{3}, x \in R$ , is the interval :

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Class 11th Chemical Equilibrium

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${(t a n^{- 1} x)}^{3} + {(c o t^{- 1} x)}^{3} = k π^{3}, x \in R$

$\Rightarrow (t a n^{- 1} x + c o t^{- 1} x) ({(t a n^{- 1} x + c o t^{- 1} x)}^{2} - 3 t a n^{- 1} x c o t^{- 1} x) = k π^{3}$

$\Rightarrow \frac{1}{3} \leq k < \frac{7}{8}$

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PCl₅ dissociates as

PCl₅(g) $?$ PCl₃ (g) + Cl₂(g)

5 moles of PCl₅ placed in a 200 litre vessel which contains 2 moles of N₂ and is maintained at 600K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl₅ is ____________ * 10^-3g. (nearest integer).

(Given R = 0.082 L atm K^-1 mol^-1; Assume ideal gas behavior)

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$\begin{array}{l} P C l_{5} (g) ? P C l_{3} (g) + C l_{2} (g) \\ t = 0 5 m o l e s 0 0 \\ \underline{- x m o l e} \underline{+ x m o l e} + \underline{x m o l e} \end{array}$

Eq.(5 – x) mole x molex mole

Total number of moles at equilibrium

$= (5 - x) + x + x + n_{N_{2}} = (7 + x)$

$K_{p} = \frac{(P_{P C l_{3}}) * {(P C l_{2})}_{C l}}{{(P_{P C l_{5}})}_{e q}} = \frac{(\frac{3}{10} * 2.46) * (\frac{3}{10} * 2.46)}{(\frac{2}{10} * 2.46)} = 1.107 a t m$

= 1107 * 103 atm

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If the sum of the squares of the reciprocals of the roots a and b of the equation $3 x^{2} + λ x - 1 = 0$ is 15, then $6 {(α^{3} + β^{3})}^{2}$ is equal to :

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$α + β = - \frac{λ}{3}, α β = - \frac{1}{3}$

$∴ \frac{1}{α^{2}} + \frac{1}{β^{2}} = 15 \Rightarrow λ = \pm 3$

$∴ 6 {(α^{3} + β^{3})}^{2} = 6 {({(α + β)}^{3} - 3 α β (α + β))}^{2} = 6 * {(- (1 + 1))}^{2} = 24$

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Class 11th Chemistry

At 300K, a sample of 3.0g of gas A occupies the same volume as 0.2g of hydrogen at 200K at the same pressure. The molar mass of gas A is _________g mol^-1, (nearest integer) Assume that the behavior of gases as ideal. (Given: The molar mass of hydrogen (H₂) gas is 2.0g mol^-1.)

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${P V)}_{H_{2}} = {(P V)}_{G a s}$

${(n R T)}_{H_{2}} = {(n R T)}_{G a s}$

$n_{H_{2}} * T_{H_{2}} = n_{G a s} * T_{G a s}$

$\Rightarrow \frac{0.2}{2} * 200 = \frac{3}{{(M M)}_{G a s}} * 300 \Rightarrow {(M M)}_{G a s} = 45 g m / m o l e$

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Let x²+ y² + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x² at (2, 4). Then A + C is equal to

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Equation of tangent to the circle at (2, 4) is

(4 + A) x + (8 + B) y + 2A + 2B + 2C = 0……. (i)

Also equation of tangent to parabola y = x² at (2, 4) is

4x – y = 4 ………. (ii)

Comparing (i) and (ii) we get, A + C = 16

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The remainder when 3²⁰²²is divided by 5 is :

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(3 * 3 * 3 * ………2022 times) ¸ 5

Remainder = (-1) (-1) (-1) ….1011 times)

Remainder = -1 + 5 = 4

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Class 11th Sets

Let A = ${z \in C : 1 \leq | z - (1 + i) \leq 2}$ and B = ${z \in A : | z - (1 - i) | = 1}$ . Then, B :

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$A = {z \in c : 1 \leq | z - (1 + i) | \leq 2}$

| z - (1 + i) | \geq 1

and $| z - (1 + i) | \leq 2$

and also $| z - (1 - i) | = 1$

hence B has infinite set.

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Let P₁ be a parabola with vertex (3, 2) and focus (4, 4) and P₂ be its mirror image with respect to the line x + 2y = 6. Then the directrix P₂ is x + 2y =.

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