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.
Step $1$: Calculate the cell constant $(G^*)$.
$G^* = \kappa \times R = (2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}) \times (164 \ \Omega) = 0.4539 \ cm^{-1}$.
Step $2$: Calculate the conductivity $(\kappa)$ of $0.05 \ M \ AgNO_3$.
$\kappa = \frac{G^*}{R} = \frac{0.4539 \ cm^{-1}}{75.8 \ \Omega} = 5.988 \times 10^{-3} \ \Omega^{-1} \ cm^{-1} \approx 5.99 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.
Step $3$: Calculate the molar conductivity $(\Lambda_m)$ of $AgNO_3$.
$\Lambda_m = \frac{1000 \times \kappa}{M} = \frac{1000 \times 5.988 \times 10^{-3}}{0.05} = 119.76 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$G^* = \kappa \times R = (2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}) \times (164 \ \Omega) = 0.4539 \ cm^{-1}$.
Step $2$: Calculate the conductivity $(\kappa)$ of $0.05 \ M \ AgNO_3$.
$\kappa = \frac{G^*}{R} = \frac{0.4539 \ cm^{-1}}{75.8 \ \Omega} = 5.988 \times 10^{-3} \ \Omega^{-1} \ cm^{-1} \approx 5.99 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.
Step $3$: Calculate the molar conductivity $(\Lambda_m)$ of $AgNO_3$.
$\Lambda_m = \frac{1000 \times \kappa}{M} = \frac{1000 \times 5.988 \times 10^{-3}}{0.05} = 119.76 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
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