How is the cell constant determined?

  • A
    By measuring the resistance of the cell containing a solution of known conductivity.
  • B
    By measuring the resistance of the cell containing a solution of unknown conductivity.
  • C
    By measuring the length and area of the electrodes.
  • D
    By measuring the potential difference across the cell.

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Similar Questions

Calculate $\Lambda _m^o$ for $CaCl_2$ and $MgSO_4$ from the data given in the table below:
Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$ Ion and $\lambda ^o / (S \ cm^2 \ mol^{-1})$
$H^{+} : 349.6$ $OH^{-} : 199.1$
$Na^{+} : 50.1$ $Cl^{-} : 76.3$
$K^{+} : 73.5$ $Br^{-} : 78.1$
$Ca^{2+} : 119.0$ $CH_3COO^{-} : 40.9$
$Mg^{2+} : 106.0$ $SO_4^{2-} : 160.0$

The metal which is the best conductor of electricity is

The conductivity of a $0.05 \, M$ solution of a weak monobasic acid is $10^{-3} \, S \, cm^{-1}$. If $\lambda_{m}^{\infty}$ for the weak acid is $500 \, S \, cm^{2} \, mol^{-1}$,calculate the $K_{a}$ of the weak monobasic acid.

The equivalent conductance of a monobasic acid at infinite dilution is $348 \ ohm^{-1} \ cm^2 \ eq^{-1}$. If the resistivity of the solution containing $15 \ g$ of the acid (molecular weight $49$) in $1 \ L$ is $18.5 \ ohm \ cm$,what is the degree of dissociation of the acid? (Answer in percentage)

The conductivity of $0.20 \ M \ KCl$ solution at $300 \ K$ is $0.0248 \ \Omega^{-1} \ cm^{-1}$. What is its molar conductivity?

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