If the molar conductivity of a solution is 1. | vedclass

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Question

(B) The relationship between molar conductivity $(\Lambda_m)$ and specific conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa

Vedclass · Accepted Answer

$1.26 	imes 10^{-3}$

Vedclass · Answer

(B) The relationship between molar conductivity $(\Lambda_m)$ and specific conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa 	imes 1000}{M}$.Rearranging for specific conductivity $(\kappa)$: $\kappa = \frac{\Lambda_m 	imes M}{1000}$.Given: $\Lambda_m = 1.26 	imes 10^{2} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $M = 0.01 \ mol \ L^{-1}$.Substituting the values: $\kappa = \frac{1.26 	imes 10^{2} 	imes 0.01}{1000}$.$\kappa = \frac{1.26 	imes 1}{1000} = 1.26 	imes 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

If the molar conductivity of a solution is $1.26 \times 10^{2} \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and its molarity is $0.01 \ M$,what will be its specific conductivity?

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