Conductivity of a solution is $1.26 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$. Calculate molar conductivity for $0.01 \ M$ solution.

The specific conductivity of a $0.01 \ M$ salt solution is $1.061 \times 10^{-4} \ S \ cm^{-1}$. The molar conductivity of this solution is:

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.

The values of $\lambda^{\infty}_m$ for $NH_4Cl$,$NaOH$,and $NaCl$ are $129.8$,$248.1$,and $126.4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively. Calculate $\lambda^{\infty}_m$ for $NH_4OH$ solution (in $\Omega^{-1} \ cm^2 \ mol^{-1}$).

Which of the following $N$ solutions of $NaCl$ has the maximum specific conductivity?

If the conductance and specific conductance of a solution are both $1$,then its cell constant would be: