The molar conductivity of a 0.04 text{ M} AB_ | vedclass

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Question

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa 	imes 10

Vedclass · Accepted Answer

$0.008 	ext{ }\Omega^{-1} 	ext{cm}^{-1}$

Vedclass · Answer

(B) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(\kappa)$ is given by the formula: $\Lambda_m = \frac{\kappa 	imes 1000}{M}$, where $M$ is the molarity.Given: $\Lambda_m = 200 	ext{ }\Omega^{-1} 	ext{cm}^2 	ext{mol}^{-1}$ and $M = 0.04 	ext{ M}$.Rearranging the formula to solve for $\kappa$: $\kappa = \frac{\Lambda_m 	imes M}{1000}$.Substituting the values: $\kappa = \frac{200 	imes 0.04}{1000}$.$\kappa = \frac{8}{1000} = 0.008 	ext{ }\Omega^{-1} 	ext{cm}^{-1}$.

The molar conductivity of a $0.04 \text{ M}$ $AB_2$ type salt solution at $300 \text{ K}$ is $200 \text{ }\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$. Find the conductivity.

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