The conductivity of water at 298 K is $5.54 \times 10^{- 8} S {cm}^{- 1}$ . If $λ_{H^{+}}^{\infty} = 349.8 S {cm}^{2} {mol}^{- 1}$ and $λ_{{OH}^{-}}^{\infty} = 197.8 S {cm}^{2}$ ${mol}^{- 1}$ , the degree of dissociation of water is

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$1.82 \times 10^{- 8}$

$1.82 \times 10^{- 9}$

$1.82 \times 10^{- 7}$

$1.82 \times 10^{- 6}$

answer is D.

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Detailed Solution

$\begin{array}{l} α = \frac{κ}{c Λ_{m}^{\infty} (H_{2} O)} = \frac{5.54 \times 10^{- 8} S {cm}^{- 1}}{(55.56 mol {dm}^{- 3}) \{(349.8 + 197.8) S {cm}^{2} {mol}^{- 1}\}} \\ = \frac{5.54 \times 10^{- 8} S {cm}^{- 1}}{(55.56 \times 10^{- 3} mol {cm}^{- 3}) (547.6 S {cm}^{2} {mol}^{- 1})} = 1.82 \times 10^{- 9} \end{array}$