1 × 10 ^ - 5 M AgNO _ 3 is added to 1L of saturated solution of...

The concentrations of the ions can be calculated as follows,

$\left[{Ag}^{+}\right] = {10}^{- 5}M$

\left[{NO}_{3}^{-}\right] = {10}^{- 5}M$</p><p>\left[{Br}^{-}\right] = \dfrac{Ksp}{\left[{Ag}^{+}\right]} = 4 . 9 \times {10}^{- 8}M$

${Λ}_{m} = \dfrac{k}{1000 \times M} = \dfrac{Specificconductance}{1000 \times Molarity}$

For ${Ag}^{+}$

$6 \times {10}^{- 3} = \dfrac{k{Ag}^{+}}{1000 \times {10}^{- 5}}$

${k}_{Ag +} = 6 \times {10}^{- 5}{Sm}^{- 1}$

$⇒ 6000 \times {10}^{- 8}{Sm}^{- 1}$

for ${Br}^{-}$

$8 \times {10}^{- 3} = \dfrac{k{Br}^{-}}{1000 \times 4 . 9 \times {10}^{- 8}}$

${k}_{Br -} = 39 . 2 \times {10}^{- 8}{Sm}^{- 1}$

for ${NO}_{3}^{-}$

$7 \times {10}^{- 3} = \dfrac{k{NO}_{3}^{-}}{1000 \times {10}^{- 5}}$

$k{NO}_{3}^{-} = 7 \times {10}^{- 5}{Sm}^{- 1}$

$= 7000 \times {10}^{- 8}{Sm}^{- 1}$

Conductivity of solution

$⇒ ( 6000 + 7000 + 39 . 2 ) \times {10}^{- 8}$

$⇒ 13039 . 2 \times {10}^{- 8}S{m}^{- 1}$