The specific conductance of a $0.1 \ N \ KCl$ solution at $23 \ ^oC$ is $0.012 \ \Omega^{-1} \ cm^{-1}$. The resistance of a cell containing the solution at the same temperature was found to be $55 \ \Omega$. The cell constant will be .............. $cm^{-1}$.

The equivalent conductance of $NaCl$ at concentration $C$ and at infinite dilution are $\lambda_C$ and $\lambda_{\infty}$ respectively. The correct relationship between $\lambda_C$ and $\lambda_{\infty}$ is given as: (Where the constant $B$ is positive)

The specific conductivity of a normal solution of $KCl$ at $25 \, ^oC$ is $0.002765 \, S \, cm^{-1}$. If the resistance of the cell is $400 \, \Omega$,what is the cell constant (in $cm^{-1}$)?

The resistance of $0.1 \ M$ weak acid $HA$ in a conductivity cell is $2 \times 10^3 \ \Omega$. The cell constant of the cell is $0.78 \ cm^{-1}$ and $\lambda_{m}^{\circ}$ of acid $HA$ is $390 \ S \ cm^2 \ mol^{-1}$. The $pH$ of the solution is

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 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.