NEET Practice Questions (MCQs) with Answers & Solutions

(b)

$$\begin{gathered} \mathrm{Given}, {\mathrm{N}}_2 + 3{\mathrm{H}}_2 \stackrel{ }{\rightleftharpoons }2{\mathrm{NH}}_3, {\mathrm{K}}_1 ...\left(\mathrm{i}\right) \\ {\mathrm{N}}_2 + {\mathrm{O}}_2 \stackrel{ }{\rightleftharpoons }2\mathrm{NO}, {\mathrm{K}}_2 ...\left(\mathrm{ii}\right) \\ {\mathrm{N}}_2 + \frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\stackrel{ }{\to }}{\mathrm{H}}_2\mathrm{O}, {\mathrm{K}}_3 ...\left(\mathrm{iii}\right) \\ \mathrm{To} \mathrm{calculate}, \\ 2{\mathrm{NH}}_3 + \frac{5}{2}{\mathrm{O}}_2 \stackrel{ \mathrm{K} }{\rightleftharpoons }2\mathrm{NO} +\hspace{0.17em}3{\mathrm{H}}_2\mathrm{O}, \mathrm{K} = ? ...\left(\mathrm{iv}\right) \end{gathered}$$

On reversing the equation (i) and multiplying the equation (iii) by 3, we get

$$\begin{gathered} 2{\mathrm{NH}}_3\stackrel{ }{\rightleftharpoons }{\mathrm{N}}_2 +\hspace{0.17em}3{\mathrm{H}}_2, \frac{1}{{\mathrm{K}}_1} ...\left(\mathrm{v}\right) \\ 3{\mathrm{H}}_2 + \frac{3}{2}{\mathrm{O}}_2 \stackrel{ }{\to } 3{\mathrm{H}}_2\mathrm{O}, {\mathrm{K}}_3^3 ...\left(\mathrm{vi}\right) \\ \mathrm{Now}, \mathrm{add} \mathrm{equation}. \left(\mathrm{ii}\right), \left(\mathrm{v}\right) \mathrm{and} \left(\mathrm{vi}\right), \mathrm{we} \mathrm{get} \mathrm{the} \mathrm{resultant} \mathrm{equation}. \left(\mathrm{iv}\right). \\ 2{\mathrm{NH}}_3 + \frac{5}{2}{\mathrm{O}}_2 \stackrel{ \mathrm{K} }{\rightleftharpoons }2\mathrm{NO} + 3{\mathrm{H}}_2\mathrm{O} \\ \therefore \mathrm{K}=\frac{{\mathrm{K}}_2{\mathrm{K}}_3^3}{{\mathrm{K}}_1} \end{gathered}$$

(b) Key Idea As solubility of AgCl(s) is asked in 0.1 M NaCl, so in the calculation, solubility of C^l- (from NaCl) must be added to the solubility of Cl^- (from AgCl).

Let s be the solubility of Ag⁺ and Cl^- in AgCl before the addition of NaCl.

         NaCl (aq) $\to$Na⁺ (aq) + Cl^-(aq)

          0.1 M          0                0

           0               0.1 M         0.1+s

       AgCl(s) $\rightleftharpoons$Ag⁺(aq) + Cl^-(aq)

                         s             s+0.1

Given,           K_sp = 1.6 x 10^-10 = [Ag⁺][Cl^-]

        or                   1.6 x 10^-10 = s(0.1+s)

                                                = 0.1s + s²

._.. K_sp is small, so s very less in comparison with 0.1. Hence s² can be neglected.

         Thus, 1.6 x 10-10 = 0.1s

   or                          s = 1.6 x 10-9 M

(c) Boric acid can be considered as an acid because its molecule accepts OH' from water, releasing proton.

H₃BO₃ + H₂O $\rightleftharpoons$   B(OH)^-₄   +   H⁺

  Acid     Base     Conjugate  Conjugate

                            base           acid

Remember In the given options to the question, (a), (b) and (c) are correct as all of these sentences have more or less similar meaning but here (c) option is the most appropriate one as it gives complete explanation of the fact that how boric acid can be combined with an acid.

Among the given options, PF3 (phosphorus trifluoride) is the most likely to behave as a Lewis base. It has an available lone pair of electrons on the phosphorus atom, which can be donated to an electron-deficient Lewis acid, forming a coordinate covalent bond.

(a) As the oxidation state of halogen i.e. —Cl in this case increases, acidity of oxyacid increases.

HClO : Oxidation state of Cl = +1

HClO₂: Oxidation state of Cl = + 3

HClO₂; Oxidation state of Cl = + 5

HClO₄ Oxidation state of Cl = + 7

Therefore, the correct order of acidity would be

HClO < HClO₂ < HClO₃ < HClO₄

(a) In the nitration of benzene in the presence of conc. H₂SO₄ and HNO₃, nitrobenzene is formed.

HNO₃ + H₂SO₄ => NO₂⁺+ HSO₄ ^-+ H₂O

                                  Electrophilc  Nucleophile

If large amount of KHSO₄ is added to this mixture, more HSO₄ ion furnishes and hence the concentration of NO₂, i.e. electrophile decreases.

As concentration of electrophile decreases, rate of electrophilic aromatic reaction slows down.

For a reaction,

$$\begin{gathered} \underset{\mathrm{Reactant}}{\mathrm{A}}\rightleftharpoons \underset{\mathrm{Product}}{\mathrm{B}} \\ \mathrm{K}=\frac{{\left[\mathrm{B}\right]}_{\mathrm{eq}}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}}}\Rightarrow 1.6\times {10}^{12}=\frac{{\left[\mathrm{B}\right]}_{\mathrm{eq}}}{{\left[\mathrm{A}\right]}_{\mathrm{eq}}} \\ {\left[\mathrm{B}\right]}_{\mathrm{eq}}>>{\left[\mathrm{A}\right]}_{\mathrm{eq}} \end{gathered}$$

So, mostly the product will be present in te equilibrium mixture.