An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

  • A
    increase in ionic mobility of ions
  • B
    $100\%$ ionisation of electrolyte at normal dilution
  • C
    increase in both $i.e.$,number of ions and ionic mobility of ions
  • D
    increase in number of ions.

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Similar Questions

The equivalent conductance at infinite dilution $\wedge^0$ for electrolytes $BA$ and $CA$ are $140$ and $120 \ S \ cm^2 \ eq^{-1}$,respectively. The equivalent conductance at infinite dilution for $BX$ is $198 \ S \ cm^2 \ eq^{-1}$. The $\wedge^0$ (in $S \ cm^2 \ eq^{-1}$) of $CX$ is:

Which is the conductivity of $0.02 \ M \ HCl$ solution if molar conductivity of the solution at $25^{\circ} C$ is $412.3 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$?

What is the ionic mobility of $Ag^{+}$? Given $\lambda^{Ag^{+}} = 5 \times 10^{-4} \ \Omega^{-1} \ cm^{2} \ eq^{-1}$.

Conductivity of $0.001 \ M$ aqueous solution of $Na_2SO_4$ is found to be $2.6 \times 10^{-3} \ S \ cm^{-1}$ at $25 \ ^oC$. If limiting molar conductance of $Na^+$ is $50 \ S \ cm^2 \ mol^{-1}$,then limiting molar conductance of $SO_4^{2-}$ will be .............. $S \ cm^2 \ mol^{-1}$ (neglect conductivity of water).

Resistance of a cell containing $0.02 \ M \ KCl$ solution is $164 \ \Omega$. If the cell is filled with $0.05 \ M \ AgNO_3$,the resistance becomes $75.8 \ \Omega$. Calculate the following: [Conductivity of $0.02 \ M \ KCl = 2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$] $(i)$ Conductivity of $0.05 \ M \ AgNO_3$ (ii) Molar conductivity of $AgNO_3$ solution.

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