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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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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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a year 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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a year 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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a year ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

Metal M crystallizes into a fcc lattice with the edge length of 4.0 * 10^-8cm. The atomic mass of the metal is_________ g/mol. (Nearest integer)

(Use: N_A = 6.02 * 10²³mol^-1, density of metal, M = 9.03 g cm^-3).

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Vishal Baghel

Contributor-Level 10

$a = 4 * 1 0^{- 8} c m z (F . C . C) = 4$

d = 9.03 g /ml

$M = \frac{d * a^{3} * N_{A}}{Z}$

$= \frac{9.03 * {(4 * 1 0^{- 8})}^{3} * 6.02 * 1 0^{23}}{4}$

$= 86.97 \approx 87$

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a year ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

2L of 0.2 M H₂SO₄ is reacted with 2L of 0.1 M NaOH solution, the molarity of the resulting product Na₂SO₄ in the solution is_________ millimolar (Nearest integer)

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Vishal Baghel

Contributor-Level 10

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

Number of mole = 0.2 * 2 2 * 0.1 - -

0.4 = 0.2

No. of moles after reaction = 0.4 – 0.1 0.2 – 0.2 &nbs

...more

H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O

Number of mole = 0.2 * 2 2 * 0.1 - -

0.4 = 0.2

No. of moles after reaction = 0.4 – 0.1 0.2 – 0.2 0.1

= 0.3 = 0

molarity of Na₂SO₄ = $\frac{n}{V (L)} = \frac{0.1}{4} = 0.025$

$∴$ = 25 milli molar

less

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a year ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight

At $20.0 m L$ solution containing $0.2 g$ impure $H_{2} O_{2}$ reacts completely with $0.316 g$ of ${K M n O}_{4}$ in acid solution. The purity of $H_{2} O_{2}$ (in %) is (mol. wt. of $H_{2} O_{2} = 34$ ; mol. wt. of ${K M n O}_{4} = 158$ )

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alok kumar singh

Contributor-Level 10

eq of $H_{2} O_{2} =$ eq of ${K M n O}_{4}$

$\begin{array}{r} = \frac{0.316}{158} * 5 = 0.01 \\ = \frac{0.01 * 17}{0.2} * 100 = 85 % \end{array}$

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a year ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Twelve

The number of chiral centres present in $[B]$ is

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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a year ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter One

The mass of ammonia in grams produced when $2.8 k g$ of dinitrogen quantitatively reacts with $1 k g$ of dihydrogen is(Mole Concept)

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alok kumar singh

Contributor-Level 10

$N_{2} + 3 H_{2} ? 2 {N H}_{3}$

$\begin{array}{l} \frac{2.8 * 10^{3}}{28} & \frac{1 * 10^{3}}{2} \\ \frac{0.1 * 10^{3} m o l}{L R} & 0.5 * 10^{3} m o l \end{array}$

$Mass of {N H}_{3} produced = 0.2 * 10^{3} * 17 = 3.4 k g = 3400 g$

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a year ago

Chemistry Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R

Assertion A: Thin layer chromatography is an adsorption chromatography.

Reason R: A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent.

In the light of the above statements, choose the correct answer from the options given below

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Vishal Baghel

Contributor-Level 10

Thin layer chromatography is another type of adsorption chromatography silica gel of proper size spread over glass plate also acts as an adsorbent in thin layer chromatography size of 0.2 mm thick.

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