Given below is the plot of the molar conducti | vedclass

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(A) From the graph,for the concentrated solution $(c)$: $\sqrt{C_c} = 0.15 \ (mol/L)^{1/2}$,so $C_c = (0.15)^2 = 0.0225 \ mol/L$. The molar conductivi

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$150$

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(A) From the graph,for the concentrated solution $(c)$: $\sqrt{C_c} = 0.15 \ (mol/L)^{1/2}$,so $C_c = (0.15)^2 = 0.0225 \ mol/L$. The molar conductivity $\Lambda_{m,c} = 100 \ S \ cm^2 \ mol^{-1}$.For the dilute solution $(d)$: $\sqrt{C_d} = 0.1 \ (mol/L)^{1/2}$,so $C_d = (0.1)^2 = 0.01 \ mol/L$. The molar conductivity $\Lambda_{m,d} = 150 \ S \ cm^2 \ mol^{-1}$.We know that $\kappa = \frac{\Lambda_m \cdot C}{1000}$ and $R = \frac{G^*}{\kappa}$,where $G^*$ is the cell constant.Thus,$R = \frac{1000 \cdot G^*}{\Lambda_m \cdot C}$.Taking the ratio of resistances: $\frac{R_d}{R_c} = \frac{\Lambda_{m,c} \cdot C_c}{\Lambda_{m,d} \cdot C_d}$.Substituting the values: $\frac{R_d}{100} = \frac{100 	imes 0.0225}{150 	imes 0.01} = \frac{2.25}{1.5} = 1.5$.Therefore,$R_d = 100 	imes 1.5 = 150 \ \Omega$.

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