A 5 . 0mmol dm ^ - 3 aqueous solution of KCl has a conductance of 0...

Given {conc}^{n} of $KCl = \dfrac{m . mol}{L}$

: Conductance \left(G\right) = 0 . 55mS$</p><p>: Cell constant \left(\dfrac{ℓ}{A}\right) = 1 . 3{cm}^{- 1}$

To Calculate : Molar conductivity \left({\lambda}_{m}\right)$of sol.</p><p>$\to$Since${Λ}_{m} = \dfrac{1}{1000} \times \dfrac{k}{M} . . . ( 1 )$</p><p>$\to$Molarity$= 5 \times {10}^{- 3}\dfrac{mol}{L}$</p><p>$\to$Conductivity$= G \times \left(\dfrac{ℓ}{A}\right) = 0 . 55mS \times \dfrac{1 . 3}{\dfrac{1}{100}}{m}^{- 1}$</p><p>$= 55 \times 1 . 3{mSm}^{- 1}$</p><p>${eq}^{n}\left(1\right) {Λ}_{m} = \dfrac{1}{1000} \times \dfrac{55 \times 1 . 3}{\left(\dfrac{5}{1000}\right)}\dfrac{{mSm}^{2}}{mol}$</p><p>$⇒ {Λ}_{m} = 14 . 3\dfrac{{mSm}^{2}}{mol}$