The resistance and conductivity of a conductivity cell containing 0.001 M KCl solution at 298 K are 1200 Ω and 1.5 × 10^-4 S cm^-1. Calculate its…

\[\kappa=\frac{1}{R}\left(\frac{l}{A}\right) \] \(\displaystyle \implies 1.5 \times 10^{-4} \mathrm{Scm}^{-1}=\frac{1}{1200} \Omega\left(\frac{l}{A}\right)\) \(\displaystyle \left(\frac{l}{A}\right)=1 \cdot 5 \times 10^{-4} \times 1200 \mathrm{~cm}^{-1}\) \(\displaystyle \left(\frac{l}{A}\right)=0 \cdot 18 \mathrm{~cm}^{-1}\) \(\displaystyle \mathbf{\Lambda}_{m}=\frac{\kappa}{C} \times 1000 \Omega^{-1} \boldsymbol{c m}^{2} \boldsymbol{m o l}^{-1}\) \(\displaystyle =\frac{1 \cdot 5 \times 10^{-4}}{0 \cdot 001} \times 1000 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) \(\displaystyle \mathbf{\Lambda}_{m}=150 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\) or \(\displaystyle 150 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)