The unit of specific conductance is:

The $k=$ $4.95$ $\times$ ${10}^{-5}S$ $c{m}^{-1}$ for a 0.00099 M solution. The reciprocal of the degree of dissociation of acetic acid, if ${\wedge }_m^0$ for acetic acid is 400 S $c{m}^{2}mo{l}^{-1}$ will be:

\(\land^o_m\) for NaCl, HCl and \(\mathrm{CH_3COONa }\) are 126.4, 425.9, and 91.05 S cm² mol^–1 respectively. If the conductivity of 0.001028 mol L^–1 acetic acid solution is \(4.95 \times 10^{-5} S ~cm^{-1} \), the degree of dissociation of the acetic acid solution is:

Limiting molar conductivities, for the given solutions, are :

${\lambda }_m^0\left(H_{2}S{O}_4\right)= x$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(K_{2}S{O}_4\right)= y$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

${\lambda }_m^0\left(C{H}_{3}COOK\right)= z$ $\mathrm{S }c{m}^2$ $mo{l}^{-1}$

From the data given above, it can be concluded that \(\lambda_m^0 \) in (\(S\ cm^2\ mol^{-1}\)) for CH₃COOH will be :
1. \(\mathrm{x-y+2z}\)       
2. \(\mathrm{x+y+z}\)          
3. \(\mathrm{x-y+z}\)       
4. \(\mathrm{{(x-y) \over 2}+z}\)          

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

1. 0.142 cm^–1
2. 0.66 cm^–1
3. 0.918 cm^–1
4. 1.12 cm^–1

The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 × 10^–3 S cm^–1 at 298 K is:
1. 11.5 S cm²/mol
2. 21.5 S cm²/mol
3. 31.5 S cm²/mol
4. 41.5 S cm²/mol

Consider the following graph.

The strong electrolyte in the above graph is represented by:

1. X

2. Y

3. Both X and Y

4. Data given is not sufficient to predict.

The cell constant of a conductivity cell-