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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Twelve

$[P]$ on treatment with ${B r}_{2} / {F e B r}_{3}$ in ${C C l}_{4}$ produced a single isomer $C_{8} H_{7} O_{2} B r$ while heating $[P]$ with sodalime gave toluene. The compound $[P]$ is:

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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11 months ago

Chemistry Aldehydes, Ketones and Carboxylic Acids

The correct structure of product 'A' formed in the following reaction, is

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Payal Gupta

Contributor-Level 10

Consider the following image

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11 months ago

Chemistry Haloalkanes and Haloarenes

The major product (P) in the reaction

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Payal Gupta

Contributor-Level 10

Consider the image below

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

Among the following the number of state variable is____________.

Internal energy (U)

Volume (V)

Heat (q)

Enthalpy (H)

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Vishal Baghel

Contributor-Level 10

State functions are

(a) Internal energy

(b) Volume

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Four

For a reaction, given below is the graph of ink vs $\frac{1}{T} .$ The activation energy for the reaction is equal to_____________ cal mol^-1 (Nearest integer)

(Given: R = 2 cal K^-1 mol^-1)

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Vishal Baghel

Contributor-Level 10

$k = A e^{- E a / R T}$

$l n k = l n A - \frac{E a}{R T}$

Comparing with y = mx + c

Slope = $+ \frac{E a}{R} = + \frac{20}{5}$

$∴ E a = \frac{20}{5} R = 4 R = 4 * 2 c a l$

= 8 Cal

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight

On reaction with stronger oxidizing agent like KlO₄,hydrogen peroxide oxidizes with the evolution of O₂. The oxidation number of I in KlO₄, changes to__________.

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Vishal Baghel

Contributor-Level 10

$I O_{4}^{-} + H_{2} O_{2} \to I O_{3}^{-} + O_{2}$

Oxidation number of l in, $I O_{4}^{-} \Rightarrow x + (- 2) * 4 = - 1 ∴ x = + 7$

Oxidation number of in, $I O_{3}^{-} \Rightarrow x + (- 2) * 3 = - 1 ∴ x = + 5$

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

A sample of 0.125g of an organic compound when analyzed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mm Hg. The percentage of nitrogen in the given organic compound is__________. (Nearest integer)

[Given: (a) The vapour pressure of water of 280K is 14.2mm Hg

(b) R = 0.082 L atm K^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

$V_{N_{2}} = 22.78 m l T = 280 K$

PTotal = 759 mm of Hg

$P_{N_{2}} = 759 - 14.2 = 744.8 m m H g$

$n_{N_{2}} = \frac{744.8 * (22.78 / 1000)}{(0.082 * 760) * 280}$

= 0.00097

$% N = \frac{0.02716}{0.125} * 100 = 21.728$

$\approx 22 .$

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

At 600K, 2 mol of NO are mixed with 1 mol of O₂.

2NO_(g) + O₂(g) $⇄$ 2NO₂(g)

The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is__________. (Nearest integer)

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Vishal Baghel

Contributor-Level 10

$2 N O_{(g)} + O_{2 (g)} ? 2 N O_{2} (g)$

Initial moles2mol1 mol

At equilibrium (2 – 2x) mole (1 – x) mol2x mol

$n o_{2} \Rightarrow (1 - x) = 0.6$

$∴ x = 0.4$

$n_{N O} = 2 - 2 x = 2 - 2 * 0.4$

= 2 0.8 = 1.2

$n_{N O_{2}} = 2 x = 2 * 0.4 = 0.8$

$∴ K p = \frac{{(\frac{0.8}{2.6})}^{2}}{{(\frac{1.2}{2.6})}^{2} (\frac{0.6}{2.6})}$

= 1.925

the nearest integer = 2.

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

A gaseous mixture of two substances A and B, under a total pressure of 0.8 atm is in equilibrium with an ideal solution. The mole fraction of substance A is 0.5 in the vapour phase and 0.2 in the liquid phase. The vapour pressure of pure liquid A is_________atm. (Nearest integer)

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Vishal Baghel

Contributor-Level 10

$P_{T} = P_{A} + P_{B} = 0.8 ? ? a t m$

$P_{A} = P_{B} = 0.4 a t m$

$Y_{A} = 0.5 Y_{B} = 0.5$

$X_{A} = 0.2$

$P_{A}^{o} = 2 a t m .$

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11 months ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

If the wavelength for an electron emitted from H-atom is 3.3 * 10^-10m, then energy adsorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is ___________times. (Nearest integer)

[Given: h = 6.626 * 10^-34 Js]

Mass of electron = 9.1 * 10^-31 kg

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Vishal Baghel

Contributor-Level 10

$λ = \frac{h}{\sqrt[]{2 m E}} E = K i n e t i c E n e r g y$

$E = \frac{h^{2}}{2 m λ^{2}}$

$= \frac{{(6.626 * 1 0^{- 34})}^{2}}{2 * 9.1 * 1 0^{- 31} * {(3.3 * 1 0^{- 10})}^{2}}$

= 2.215 * 10^-18

Eabsorbed – Erequired + K. E

$\frac{E a b s o r b e d}{E r e q u i r e d} = 1 + \frac{K . E}{E r e q u i r e d}$

= $= 1 + \frac{2.215 * 1 0^{- 18}}{13.6 * 1.602 * 1 0^{- 19}} = 2.016$

$\approx$ 2

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