The molar conductivities $\Lambda_{NaOH}^o$ and $\Lambda_{HCl}^o$ at infinite dilution in water at $25 ^\circ C$ are $91.0$ and $426.2 \ S \ cm^2/mol$ respectively. To calculate $\Lambda_{CH_3COOH}^o$,the additional value required is:

Resistivity of $0.5 \ M$ electrolytic solution is $10 \ \Omega \ cm$. Find the molar conductivity of the solution.

Molar ionic conductivities of a divalent cation and a divalent anion are $57 \ S \ cm^2 \ mol^{-1}$ and $73 \ S \ cm^2 \ mol^{-1}$ respectively. The molar conductivity of a solution of an electrolyte containing these ions will be:

If the values of $\Lambda_{\infty}$ of $NH_4Cl$,$NaOH$ and $NaCl$ are $130$,$217$ and $109 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ respectively,the $\Lambda_{\infty}$ of $NH_4OH$ in $ohm^{-1} \ cm^2 \ equiv^{-1}$ is:

Which is the conductivity of $0.02 \ M \ HCl$ solution if molar conductivity of the solution at $25^{\circ} C$ is $412.3 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$?

At $298 \ K$,the limiting molar conductivity of a weak monobasic acid is $4 \times 10^2 \ S \ cm^2 \ mol^{-1}$. At $298 \ K$,for an aqueous solution of the acid,the degree of dissociation is $\alpha$ and the molar conductivity is $y \times 10^2 \ S \ cm^2 \ mol^{-1}$. At $298 \ K$,upon $20$ times dilution with water,the molar conductivity of the solution becomes $3y \times 10^2 \ S \ cm^2 \ mol^{-1}$. $(1)$ The value of $\alpha$ is. . . . . . $(2)$ The value of $y$ is. . . . . .