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Chemistry NCERT Exemplar Solutions Class 12th Chapter One

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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One General Principles & Processes of Isolating Elemen

Match List I with List II:

List-I (Processes/Reactions)	List-II (Catalyst)
(A) 2SO₂(g) + O₂(g) -->2SO₃(g)	(I) Fe(s)
(B) 4NH₃(g) + 5O₂(g)--> 4NO(g) + 6H₂O(g)	(II) Pt(s) – Rh(s)
(C) N₂(g) + 3H₂(g) --> 2NH₃(g)	(III) V₂O₅
(D) Vegetable oil (I) + H₂ --> Vegetable ghee(s)	(IV) Ni(S)

Choose the correct answer from the options given below:

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Vishal Baghel

Contributor-Level 10

$2 S O_{2} (g) + O_{2} (g) \to_{}^{V_{2} O_{5}} 2 S O_{3} (g) c o n t a c t p r o c e s s$

$4 N H_{3} (g) + 5 O_{2} (g) \to_{}^{A (s) - R h (s)} 4 N O (g) + 6 H_{2} O (g)$ ostwald's process

$N_{2} (g) + 3 H_{2} (g) \to_{}^{F e (s)} 2 N H_{3} (g) H a b e r' s p r o c e s s$

Vegetable oil (l) + H₂ $\to_{}^{N i (s)}$ vegetable ghee; Hydrogenation

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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Chemistry Chemical Bonding and Molecular Structure

Match List I with List II:

Choose the correct answer form the options given below:

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Vishal Baghel

Contributor-Level 10

BrF₅ - Square pyramid

[CrF₆]³^- - Octahedral

O₃ - Bent

PCl₅ - Trigonal Bipyramid

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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Chemistry Some Basic Concepts of Chemistry

Hemoglobin contains 0.34% of iron by mass. The number of Fe atoms in 3.3g of hemoglobin is (Given: Atomic mass of Fe is 56 u, N_A = 6.022 * 10²³ mol^-1.)

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alok kumar singh

Contributor-Level 10

Number of Fe atoms = $\frac{0.34}{100} * \frac{3.3}{56} * 6.022 * 1 0^{23}$

= 1.206 * 10²⁰

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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Some Basic Concepts of Chemistry

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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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Class 12th

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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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Some Basic Concepts of Chemistry

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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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Some Basic Concepts of Chemistry

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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2 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Solid States

The incorrect statement about the imperfections in solids is

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Payal Gupta

Contributor-Level 10

· Density of solid decreases in Schottky defect and vacancy defect.

· Density of solid increases in interstitial defect.

· Density of solid remain unchanged in Frenkel defect.

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3 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Class 12th

Ionic radii of cation A+ and anion B are 102 and 181 pm respectively. These ions are allowed to crystallize into an ionic solid. This crystal has cubic close packing for B. A+ is present in all octahedral voids. The edge length of the unit cell of the crystal AB is________ pm.

(Nearest integer)

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Vishal Baghel

Contributor-Level 10

Edge length of the unit cell AB = a = 2 $(r_{+} + r_{-})$

$∴ a = 2 (102 + 181) = 2 (283) = 566$

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New answer posted

3 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter One Some Basic Concepts of Chemistry

$N_{2 (g)} + 3 H_{2 (g)} ? 2 N H_{3 (g)}$

20g 5g

Consider the above reaction, the limiting reagent of the reaction and number of moles of NH₃ formed respectively are:

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Vishal Baghel

Contributor-Level 10

t = 0 $\frac{20}{28}$ $\frac{5}{2}$ 0

Here, N₂ is the limiting reagent

and No of moles of NH₃ = 2 * No of moles ofN₂ reacted

$2 * \frac{20}{28} = \frac{40}{28} = \frac{1.42}{}$

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