Ncert Solutions Chemistry Class 11th

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

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)

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

Mole of CH? = 6.4 / 16 = 0.4 and mole of CO? = 8.8 / 44 = 0.2
Total mole = (0.4 + 0.2) = 0.6 mole of a non-reacting mixture of gas
Using Ideal Gas Law; P = nRT / V
P = (0.6 * 8.314 * 300) / 10 = 149.65 kPa
Ans = 150 (Rounded off)

Partial Pressure of O? = K? * solubility (K? = Henry's constant)
Solubility = PO? / K? = 20 / (8.0 * 10? ) = 2.5 * 10? = 25 * 10? M
Ans = 25

In an aqueous medium, all alkali metal cations are converted into their hydrated form. More intensely positive cations have more hydration, and hence the size of the cation increases after hydration. Therefore, the Li? cation has a larger size in hydrated form and the least conductivity. The order of conductivity of alkali metal ions is: Li? < Na? < K? < Rb? < Cs?

Except for option (D), all are the important characteristics of D? O. D? O is useful for the isotopic leveling effect in reaction mechanisms. The dielectric constants of H? O and D? O are respectively 80 and 60.

Excess of nitrogen and phosphorus is primarily responsible for eutrophication and hence an indicator of polluted environment.

In water gas shift reaction carbon monoxide is oxidized into carbon dioxide by treating it with steam in presence of catalyst.

Na? SO? + H? SO? → SO? + Na? SO? + H? O
SO? + 2NaOH → Na? SO? + H? O
Na? SO? + SO? + H? O → 2NaHSO?