The molar conductivity is maximum for the solution of concentration (in $M$)

Conductivity of a conductor is

$\Lambda_{m(HAc)}^0$ is equal to . . . . . . .

Conductivity of a saturated solution of a sparingly soluble salt $AB$ at $298 \ K$ is $1.85 \times 10^{-5} \ S \ m^{-1}$. Solubility product of the salt $AB$ at $298 \ K$ is. Given $\Lambda_{m}^{\circ}(AB) = 140 \times 10^{-4} \ S \ m^{2} \ mol^{-1}$.

Molar conductivities $\left(\Lambda_{m}^{\circ}\right)$ at infinite dilution of $NaCl$,$HCl$,and $CH_{3}COONa$ are $126.4$,$425.9$,and $91.0 \ S \ cm^{2} \ mol^{-1}$ respectively. $\Lambda_{m}^{\circ}$ for $CH_{3}COOH$ will be $:-$

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 $74.5 \; g \; mol^{-1}$.