At a particular temperature,the ratio of molar conductance to specific conductance of $0.01 \ M$ $NaCl$ solution is

The unit of molar conductivity is

Given the following molar conductivities at $25\,\text{}^{\circ}C$: $HCl = 426\,\Omega^{-1}\,cm^{2}\,mol^{-1}$,$NaCl = 126\,\Omega^{-1}\,cm^{2}\,mol^{-1}$,and sodium crotonate $(NaC)$ = $83\,\Omega^{-1}\,cm^{2}\,mol^{-1}$. Calculate the ionization constant $(K_{a})$ of crotonic acid $(HC)$ if the conductivity of a $0.001\,M$ crotonic acid solution is $3.83 \times 10^{-5}\,\Omega^{-1}\,cm^{-1}$.

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 .............. $cm^{-1}$.

$A$ conductivity cell is filled with $0.01 \ M \ KCl$ and gives a resistance of $484 \ \Omega$. The conductivity of the solution is $0.00141 \ \Omega^{-1} \ cm^{-1}$ at $25^{\circ}C$. What is the cell constant (in $cm^{-1}$)?

Correct order of limiting molar conductivity for cations in water at $298 \ K$ is $:$