$\wedge _m^o$ for $NaCl$,$HCl$ and $NaA$ are $126.4$,$425.9$ and $100.5 \ S \ cm^2 \ mol^{-1}$,respectively. If the conductivity of $0.001 \ M \ HA$ is $5 \times 10^{-5} \ S \ cm^{-1}$,degree of dissociation of $HA$ is

The specific conductance of a solution depends upon -

The molar conductivity of $KCl$ solutions at different concentrations at $298 \ K$ is given below:

$c / mol \ L^{-1}$	$\Lambda_m / S \ cm^2 \ mol^{-1}$
$0.000198$	$148.61$
$0.000309$	$148.29$
$0.000521$	$147.81$
$0.000989$	$147.09$

Show that a plot of $\Lambda_m$ versus $c^{1/2}$ is a straight line. Determine the values of $\Lambda_m^o$ and $A$ for $KCl$.

When a certain conductivity cell was filled with $0.1 \ M \ KCl$,it had a resistance of $85 \ \Omega$ at $25 \ ^oC$. When the same cell was filled with an aqueous solution of $0.052 \ M$ unknown electrolyte,the resistance was $96 \ \Omega$. Calculate the molar conductivity of the unknown electrolyte at this concentration ............. $\Omega^{-1} \ cm^2 \ mol^{-1}$ (Given: Specific conductance of $0.1 \ M \ KCl = 1.29 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$)

The conductivity of a solution of concentration $0.1 \ mol \ L^{-1}$ of a weak monobasic acid $(HA)$ (in $S \ cm^{-1}$) is (Given $\Lambda_{HA}^{\circ}=400 \ S \ cm^2 \ mol^{-1}$ and degree of dissociation $(\alpha)$ of $HA=0.02$)

If $0.01 \, M$ solution of an electrolyte has a resistance of $40 \, \Omega$ in a cell having a cell constant of $0.4 \, cm^{-1}$,then its molar conductance in $ohm^{-1} \, cm^2 \, mol^{-1}$ will be :-