Match the column $I$ with column $II$ and mark the appropriate choice.
Column $I$Column $II$
$A$. Kohlrausch law$i$. $\Lambda _{m}^o = \nu _+ \lambda _+^o + \nu _- \lambda _-^o$
$B$. Molar Conductivity$ii$. $\Lambda _m = \frac{\kappa \times 1000}{M}$
$C$. Degree of Dissociation$iii$. $\alpha = \frac{\Lambda _m}{\Lambda _m^o}$
$D$. Dissociation Constant$iv$. $K_a = \frac{C\alpha ^2}{1 - \alpha}$

  • A
    $A \to (iii), B \to (iv), C \to (i), D \to (ii)$
  • B
    $A \to (i), B \to (ii), C \to (iii), D \to (iv)$
  • C
    $A \to (iv), B \to (i), C \to (ii), D \to (iii)$
  • D
    $A \to (ii), B \to (iii), C \to (iv), D \to (i)$

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Similar Questions

Match List-$I$ with List-$II$.
List-$I$ List-$II$
$A$. $Cd_{(s)} + 2 Ni(OH)_{3(s)} \rightarrow CdO_{(s)} + 2 Ni(OH)_{2(s)} + H_2O_{(l)}$ $I$. Primary battery
$B$. $Zn(Hg) + HgO_{(s)} \rightarrow ZnO_{(s)} + Hg_{(l)}$ $II$. Discharging of secondary battery
$C$. $2 PbSO_{4(s)} + 2 H_2O_{(l)} \rightarrow Pb_{(s)} + PbO_{2(s)} + 2 H_2SO_{4(aq)}$ $III$. Fuel cell
$D$. $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$ $IV$. Charging of secondary battery

Choose the correct answer from the options given below.

Why is electrochemistry important for the invention of the latest technology?

During the electrolysis of concentrated $H_2SO_4$,perdisulphuric acid $(H_2S_2O_8)$ and $O_2$ are formed at the anode in equimolar amounts. The moles of $H_2$ that will form simultaneously at the other electrode will be (Given: $2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-$)

For a spontaneous reaction,determine the values of $\Delta G^o$,equilibrium constant $K$,and $E^o_{cell}$ respectively.

Copper reduces $NO_{3}^{-}$ into $NO$ and $NO_{2}$ depending upon the concentration of $HNO_{3}$ in solution. Assuming fixed $[Cu^{2+}]$ and $P_{NO} = P_{NO_{2}} = 1 \ bar$,the $HNO_{3}$ concentration at which the thermodynamic tendency for reduction of $NO_{3}^{-}$ into $NO$ and $NO_{2}$ by copper is the same is $10^{x} \ M$. The value of $2x$ is ...... .
$[Given: E_{Cu^{2+} / Cu}^{\circ} = 0.34 \ V, E_{NO_{3}^{-} / NO}^{\circ} = 0.96 \ V, E_{NO_{3}^{-} / NO_{2}}^{\circ} = 0.79 \ V$ and at $298 \ K, \frac{RT}{F}(2.303) = 0.059]$

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