Resistance of a conductivity cell (cell const | vedclass

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Question

(A) Cell constant $G^* = \frac{\ell}{A} = 129 \; m^{-1} = 1.29 \; cm^{-1}$.Conductivity $\kappa = \frac{G^*}{R}$.For solution $1$ $(74.5 \; ppm)$: $\k

Vedclass · Accepted Answer

$1000$

Vedclass · Answer

(A) Cell constant $G^* = \frac{\ell}{A} = 129 \; m^{-1} = 1.29 \; cm^{-1}$.Conductivity $\kappa = \frac{G^*}{R}$.For solution $1$ $(74.5 \; ppm)$: $\kappa_1 = \frac{1.29}{100} \; S \; cm^{-1}$.Concentration $C_1 \propto 74.5 \; ppm$.For solution $2$ $(149 \; ppm)$: $\kappa_2 = \frac{1.29}{50} \; S \; cm^{-1}$.Concentration $C_2 \propto 149 \; ppm$.Since $ppm$ is proportional to molarity $(M)$,$\frac{C_1}{C_2} = \frac{74.5}{149} = \frac{1}{2}$.Molar conductivity $\wedge_m = \frac{1000 \kappa}{C}$.$\frac{\wedge_1}{\wedge_2} = \frac{\kappa_1}{\kappa_2} 	imes \frac{C_2}{C_1} = \frac{1.29/100}{1.29/50} 	imes \frac{149}{74.5} = \frac{50}{100} 	imes 2 = 1$.Given $\frac{\wedge_1}{\wedge_2} = x 	imes 10^{-3} = 1$.Therefore,$x = 1000$.

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