The molar conductivity of 0.02 M AgI at 298 | vedclass

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Question

(C) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{1000 	imes \kap

Vedclass · Accepted Answer

$2.85 	imes 10^{-3} \ \Omega^{-1} \ cm^{-1}$

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(C) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{1000 	imes \kappa}{c}$ Rearranging the formula to solve for conductivity $(\kappa)$: $\kappa = \frac{\Lambda_m 	imes c}{1000}$ Given: $\Lambda_m = 142.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ $c = 0.02 \ M = 0.02 \ mol \ L^{-1}$ Substituting the values: $\kappa = \frac{142.3 	imes 0.02}{1000} \ \Omega^{-1} \ cm^{-1}$ $\kappa = 2.846 	imes 10^{-3} \ \Omega^{-1} \ cm^{-1}$ Rounding to two decimal places,we get: $\kappa \approx 2.85 	imes 10^{-3} \ \Omega^{-1} \ cm^{-1}$

The molar conductivity of $0.02 \ M$ $AgI$ at $298 \ K$ is $142.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is its conductivity?

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