The specific conductivity of a $0.01 \ M$ salt solution is $1.061 \times 10^{-4} \ S \ cm^{-1}$. The molar conductivity of this solution is:

  • A
    $1.061 \times 10^{-4}$
  • B
    $1.061$
  • C
    $10.61$
  • D
    $106.1$

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At a certain temperature and at infinite dilution,the equivalent conductances of sodium benzoate,hydrochloric acid,and sodium chloride are $240$,$349$,and $229 \ \Omega^{-1} \ cm^2 \ equiv^{-1}$ respectively. The equivalent conductance of benzoic acid in $\Omega^{-1} \ cm^2 \ equiv^{-1}$ at the same conditions is

Molar conductivity of a weak acid $HQ$ of concentration $0.18 \ M$ was found to be $1/30$ of the molar conductivity of another weak acid $HZ$ with concentration of $0.02 \ M$. If $\lambda_{Q^{-}}^0 = \lambda_{Z^{-}}^0$, then the difference of the $pK_a$ values of the two weak acids $(pK_a(HQ) - pK_a(HZ))$ is . . . . . . (Nearest integer).
[Given: degree of dissociation $(\alpha)$ $\ll 1$ for both weak acids, $\lambda^0$: limiting molar conductivity of ions]

Explain the measurement and calculation of the resistivity of an electrolytic solution.

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The resistance of a $0.5 \, M$ solution of an electrolyte in a cell was found to be $50 \, \Omega$. If the electrodes in the cell are $2.2 \, cm$ apart and have an area of $4.4 \, cm^2$,then the molar conductivity (in $S \, m^2 \, mol^{-1}$) of the solution is:

The unit $ohm^{-1}$ is used for

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