Electrical conductance due to all the ions in $1 \ cm^3$ of a given solution is called as

Resistance of a conductivity cell (cell constant $129 \; m^{-1}$) filled with $74.5 \; ppm$ solution of $KCl$ is $100 \; \Omega$ (labelled as solution $1$). When the same cell is filled with $KCl$ solution of $149 \; ppm$,the resistance is $50 \; \Omega$ (labelled as solution $2$). The ratio of molar conductivity of solution $1$ and solution $2$ is i.e.,$\frac{\wedge_{1}}{\wedge_{2}} = x \times 10^{-3}$. The value of $x$ is (Nearest integer). Given,molar mass of $KCl$ is $74.5 \; g \; mol^{-1}$.

Molar conductivities $(\Lambda ^o_m)$ at infinite dilution of $NaCl$,$HCl$ and $CH_3COONa$ are $126.4$,$425.9$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. $(\Lambda ^o_m)$ for $CH_3COOH$ will be .......... $S \ cm^2 \ mol^{-1}$.

Electrolytic conduction differs from metallic conduction in that in the case of electrolytic conduction:

The metal which is the best conductor of electricity is

For a strong electrolyte,the molar conductivity is given by $\wedge _m = \wedge _m^o - 10 \sqrt{C}$. Calculate $\wedge _m$ at $0.01 \ M$ if its value at $0.16 \ M$ is $200 \ S \ cm^2 \ mol^{-1}$.