Ncert Solutions Chemistry Class 11th

Match List - I with List - II:
List - I (Chemicals)
(a) Alcoholic potassium hydroxide
(b) Pd/BaSO?
(c) BHC (Benzene hexachloride)
(d) Polyacetylene

List - II (Use / Preparation / Constituent)
(i) electrodes in batteries
(ii) obtained by addition reaction
(iii) used for β-elimination reaction
(iv) Lindlar's Catalyst

Choose the most appropriate match:

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Raj Pandey

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o Alcoholic potassium hydroxide (Alc KOH):- used for β - elimination.

o Pd / BaSO? : - Lindlar's Catalyst.

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The set of elements that differ in mutual relationship from those of the other sets is:

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A diagonal relationship is observed in the periodic table between elements of period (2) and period (3).

Li and Mg

Be and Al

B and Si

Li and Na do not show a diagonal relationship.

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Fructose is an example of:

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The common positive oxidation states for an element with atomic number 24, are:

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The electron configuration for Cr (24) is [Ar] 4s¹3d?

Chromium shows common oxidation states from +2 to +6.

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Considering the below reaction, X and Y respectively are

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Raj Pandey

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Please consider the following Image

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In the ground state of atomic Fe(Z = 26), the spin- only magnetic moment is ___*10?¹ BM.
(Round off to the nearest integer).
[Given √3 = 1.73, √2 = 1.41]

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Ncert Solutions Chemistry Class 11th Amines

Contributor-Level 10

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In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _______ % (Round off to the nearest integer).
(Given atomic mass : C: 12.0u, H:1.0u, O:16.0u, N:14.0u)

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Ncert Solutions Chemistry Class 11th Solutions

Contributor-Level 10

Moles of benzene = 3.9 g / 78 g/mol . This would produce (3.9 / 78) moles of nitrobenzene in 100% conversion.
Produced moles of nitrobenzene = 4.92 g / 123 g/mol .
% yield = [ (4.92 / 123) / (3.9 / 78) ] * 100 = [ (4.92 * 100 * 78) / (123 * 3.9) ] = 80.0%
Ans = 80

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The mole fraction of a solute in a 100 molal aqueous solution is _______ *10^-2. (Round off to the nearest integer).
[Given : Atomic masses : H : 1.0u, O : 16.0u]

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Molality = (mole of solute * 1000) / wt of solvent (gm)
100 = (n_solute * 1000) / [ (1 - n_solute) * 18]
(1 - n_solute) / n_solute = 1000 / (100 * 18) = 10/18
18 (1 - n_solute) = 10 n_solute
18 - 18 n_solute = 10 n_solute
18 = 28 n_solute
n_solute = 18 / 28? 0.6428 = 64.28 * 10? ²
Ans = 64 (Rounded off)

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0.01 moles of a weak acid HA (Ka = 2.0 *10^-6) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is _______ *10^-5. (Round off to the nearest integer).
[Neglect volume change on adding HA. Assume degree of dissociation <<1]

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Contributor-Level 10

For the weak acid HA in the presence of strong acid HCl:
Ka = [ (Cα + 0.1) * Cα] / [C (1-α)] ≈ (0.1 * 10? ²α) / 10? ² = 0.1α
Given Ka = 2 * 10?
2 * 10? = 10? ¹ * α
α = 2 * 10?
Ans = 2

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The pressure exerted by a non- reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27°C is _______ kPa. (Round off to the nearest integer).
[ Assume gases are ideal, R = 8.314 J mol^-1 K^-1 Atomic masses: C: 12.0 u, H: 1.0 u, O: 16.0 u]

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