What is the conductivity of $0.05 \ M$ $NaOH$ solution having resistance $31.5 \ \Omega$ and cell constant $0.315 \ cm^{-1}$?

Resistance of a cell containing $0.02 \ M \ KCl$ solution is $164 \ \Omega$. If the cell is filled with $0.05 \ M \ AgNO_3$,the resistance becomes $75.8 \ \Omega$. Calculate the following: [Conductivity of $0.02 \ M \ KCl = 2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$] $(i)$ Conductivity of $0.05 \ M \ AgNO_3$ (ii) Molar conductivity of $AgNO_3$ solution.

If $l = \text{length}$,$R = \text{Resistance}$,and $A = \text{Area of cross-section}$,then which of the following relations is correct?

The molar conductivity is maximum for the solution of concentration $...........$ $M$.

Limiting molar conductivity of $CH_3COOH$ (i.e.,$\Lambda^o_{m(CH_3COOH)}$) is equal to:

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration $0.1 \, M$ is $100 \, \Omega .$ The conductivity of this solution is $1.29 \, S \, m^{-1}.$ Resistance of the same cell when filled with $0.2 \, M$ of the same solution is $520 \, \Omega .$ The molar conductivity of $0.2 \, M$ solution of electrolyte will be..........$\times 10^{-4} \, S \, m^2 \, mol^{-1}$