What must be the molarity of BaCl_2 solution | vedclass

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Question

(C) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(M)$ is given by: $\Lambda_m = \frac{\kappa 	im

Vedclass · Accepted Answer

$0.05$

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(C) The relationship between molar conductivity $(\Lambda_m)$, conductivity $(\kappa)$, and molarity $(M)$ is given by: $\Lambda_m = \frac{\kappa 	imes 1000}{M}$.Given: $\Lambda_m = 240 	ext{ }\Omega^{-1}	ext{cm}^2	ext{mol}^{-1}$, $\kappa = 0.012 	ext{ }\Omega^{-1}	ext{cm}^{-1}$.Rearranging the formula for $M$: $M = \frac{\kappa 	imes 1000}{\Lambda_m}$.Substituting the values: $M = \frac{0.012 	imes 1000}{240}$.$M = \frac{12}{240} = \frac{1}{20} = 0.05 	ext{ M}$.

What must be the molarity of $BaCl_2$ solution to have molar conductivity $240 \text{ }\Omega^{-1}\text{cm}^2\text{mol}^{-1}$ and conductivity $0.012 \text{ }\Omega^{-1}\text{cm}^{-1}$ at $25 \text{ }^\circ\text{C}$ (in M)?

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