The resistance of a 0.2 , M electrolyte solut | vedclass

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(B) $1$. First,calculate the cell constant $(G^*)$ using the data for the $0.2 \, M$ solution: $G^* = \kappa 	imes R = 1.3 \, S \, m^{-1} 	imes 50 \

Vedclass · Accepted Answer

$6.25 	imes 10^{-4} \, S \, m^2 \, mol^{-1}$

Vedclass · Answer

(B) $1$. First,calculate the cell constant $(G^*)$ using the data for the $0.2 \, M$ solution: $G^* = \kappa 	imes R = 1.3 \, S \, m^{-1} 	imes 50 \, \Omega = 65 \, m^{-1}$. $2$. Now,calculate the conductivity $(\kappa)$ for the $0.4 \, M$ solution using the same cell constant: $\kappa = \frac{G^*}{R} = \frac{65 \, m^{-1}}{260 \, \Omega} = 0.25 \, S \, m^{-1}$. $3$. Calculate the molar conductivity $(\Lambda_m)$ using the formula $\Lambda_m = \frac{\kappa}{C}$: Given $C = 0.4 \, M = 0.4 	imes 10^3 \, mol \, m^{-3} = 400 \, mol \, m^{-3}$. $\Lambda_m = \frac{0.25 \, S \, m^{-1}}{400 \, mol \, m^{-3}} = 6.25 	imes 10^{-4} \, S \, m^2 \, mol^{-1}$.

The resistance of a $0.2 \, M$ electrolyte solution is $50 \, \Omega$. The specific conductivity of the solution is $1.3 \, S \, m^{-1}$. If the resistance of a $0.4 \, M$ solution of the same electrolyte is $260 \, \Omega$,what is its molar conductivity?

Similar Questions

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