Chemistry MCQs for NEET — Practice Questions with Answers

(d) Key Idea The reaction for which the number of moles of gaseous products(n_p) is not equal to the number of moles of gaseous reactants (n_R), has different value of K_c and K_p.

From the equation, ${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}\times (\mathrm{RT}{)}^{∆{\mathrm{n}}_{\mathrm{g}}}$

where, $\left[∆{\mathrm{n}}_{\mathrm{g}}\mathrm{gaseous} = {\mathrm{n}}_{\mathrm{p}}-{\mathrm{n}}_{\mathrm{R}}\right]$$$

(a) n_p = n_R = 2, thus, K_p=K_c

(b) n_p = n_R = 2, thus, K_p=K_c

(c) n_p = n_R = 2, thus, K_p=K_c

(d) n_p =2, n_R = 1, thus, K_p$\ne$$$K_c

(b) K_{1 = $\frac{[\mathrm{NO}{]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}$ ;}   

      _{$K_{\mathit{2}}$= $\frac{\left[\mathrm{NO}\right]}{\left[{\mathrm{N}}_2{]}^{1/2}\right[{\mathrm{O}}_2{]}^{1/2}}$;}

        _{.^..  $K_{\mathit{1}}\mathit{=}{K}_{\mathit{2}}^{\mathit{2}}$}

(a) Ostwald's dilution law is valid only for weak electrolytes.

(d) Aspirine a weak acid is unionised in acid medium due to common ion effect and completely ionised in alkaline medium.

(d) CH₃COOH (weak acid) and CH₃COONa (conjugated salt) form acidic buffer and for acidic buffer,

         pH =pK_a + log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$ and [H⁺] = -antilog pH

          pH = -logK_a +  log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$     [^..^. pK_a = -log K_a]

               = -log(1.8 x 10-5) + log$\frac{(0.20)}{(0.10)}$

               = 4.74 + log2

               = 4.74+0.3010 =5.041

   Now,  [H⁺] = antilog(-5.045)

                    = 9.0x10^-6 mol/L

(b) e.g. , NH₄Cl ; hydrolysis of ${\mathrm{NH}}_4^{+}$ ions, i.e., cation occurs.

(b) Quotient Q=[C]³/[A][B] = (4³ x 3 x 3)/(3³ x 1 x2) = 10.66

      [(C) = 4/3 ; (A) =1/3 ; (B) =2/3]

     Since, K_c =10, Thus, Q must decrease to attain value of Kc and therefore, [C] must decrease or [A] or [B] must increase, i.e, backward direction.

(c) K₁ = $\frac{\left[{\mathrm{XeOF}}_4\right][\mathrm{HF}{]}^2}{\left[{\mathrm{XeF}}_6\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}$                                                                   ...(i)

     K₂ = $\frac{\left[{\mathrm{XeOF}}_4\right]\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]}{\left[{\mathrm{XeO}}_4\right]\left[{\mathrm{XeF}}_6\right]}$                                                             ....(ii)

  By Eq. (ii)/(i) we have K₂/K₁ = $\frac{\left[{\mathrm{XeO}}_3{\mathrm{F}}_2\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}{\left[{\mathrm{XeO}}_4\right][\mathrm{HF}{]}^2}$ = K_c

(b) Higher is pH, lesser is acidic nature. Also NH₄Cl (aq) is acidic and NaCN(aq) is alkaline. NaCl(aq) is neutral.

(b) Find solubility for each separately by s² = K_sp for MnS and ZnS,

108 S⁵ = K_sp for Bi₂S₃ and

4s³ = K_sp for Ag₂S.