The resistance of a conductivity cell filled | vedclass

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(A) Step $1$: Calculate the cell constant $(G^* = \frac{\ell}{A})$.Given for $0.1 \ M$ solution: $R = 100 \ \Omega$,$\kappa = 1.29 \ S \ m^{-1}$.$G^*

Vedclass · Accepted Answer

$124 	imes 10^{-4} \ S \ m^2 \ mol^{-1}$

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(A) Step $1$: Calculate the cell constant $(G^* = \frac{\ell}{A})$.Given for $0.1 \ M$ solution: $R = 100 \ \Omega$,$\kappa = 1.29 \ S \ m^{-1}$.$G^* = \kappa 	imes R = 1.29 \ S \ m^{-1} 	imes 100 \ \Omega = 129 \ m^{-1}$.Step $2$: Calculate conductivity $(\kappa)$ for $0.02 \ M$ solution.Given: $R = 520 \ \Omega$,$G^* = 129 \ m^{-1}$.$\kappa = \frac{G^*}{R} = \frac{129}{520} \approx 0.248 \ S \ m^{-1}$.Step $3$: Calculate molar conductivity $(\Lambda_m)$.$\Lambda_m = \frac{\kappa}{C} = \frac{0.248 \ S \ m^{-1}}{0.02 \ mol \ L^{-1}} = \frac{0.248 \ S \ m^{-1}}{0.02 	imes 10^3 \ mol \ m^{-3}} = \frac{0.248}{20} \ S \ m^2 \ mol^{-1} = 0.0124 \ S \ m^2 \ mol^{-1}$.Converting to scientific notation: $124 	imes 10^{-4} \ S \ m^2 \ mol^{-1}$.

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