Resistance of a conductivity cell at $298 \ K$ for $0.0100 \ M \ KCl$ is $161.8 \ \Omega$. Resistance becomes $190 \ \Omega$ when $0.005 \ M \ NaOH$ solution is filled in this cell. Calculate: $(i)$ cell constant,$(ii)$ specific conductivity for $NaOH$ solution,and $(iii)$ molar conductivity.
(N/A) Given: For $0.0100 \ M \ KCl$,$\kappa = 0.00141 \ S \ cm^{-1}$ and $R = 161.8 \ \Omega$.
$(i)$ Cell constant $(G^*)$ = $\kappa \times R = 0.00141 \ S \ cm^{-1} \times 161.8 \ \Omega = 0.2281 \ cm^{-1}$.
$(ii)$ For $0.005 \ M \ NaOH$,$R = 190 \ \Omega$. Specific conductivity $(\kappa)$ = $G^* / R = 0.2281 \ cm^{-1} / 190 \ \Omega = 1.2 \times 10^{-3} \ S \ cm^{-1}$.
$(iii)$ Molar conductivity $(\Lambda_m)$ = $(\kappa \times 1000) / M = (1.2 \times 10^{-3} \ S \ cm^{-1} \times 1000) / 0.005 \ M = 240 \ S \ cm^2 \ mol^{-1}$.
$(i)$ Cell constant $(G^*)$ = $\kappa \times R = 0.00141 \ S \ cm^{-1} \times 161.8 \ \Omega = 0.2281 \ cm^{-1}$.
$(ii)$ For $0.005 \ M \ NaOH$,$R = 190 \ \Omega$. Specific conductivity $(\kappa)$ = $G^* / R = 0.2281 \ cm^{-1} / 190 \ \Omega = 1.2 \times 10^{-3} \ S \ cm^{-1}$.
$(iii)$ Molar conductivity $(\Lambda_m)$ = $(\kappa \times 1000) / M = (1.2 \times 10^{-3} \ S \ cm^{-1} \times 1000) / 0.005 \ M = 240 \ S \ cm^2 \ mol^{-1}$.
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