The conductivity of water at $$298$$ K is $$5.54$$×$$10$$−$$8Scm$$−$$1$$. If λ$$H+$$∞$$=349.8Scm2mol$$−$$1$$ and λOH−∞$$=197.8Scm2$$ mol−$$1$$, the degree of dissociation of water is

α$$=$$κcΛm∞$$H2O=5.54$$×$$10$$−$$8$$ S cm−$$155.56moldm$$−$$3(349.8+197.8)S$$ $$cm2$$ mol−$$1=5.54$$×$$10$$−$$8Scm$$−$$155.56$$×$$10$$−$$3molcm$$−$$3547.6S$$ $$cm2$$ mol−$$1=1.82$$×$$10$$−$$9$$

Questions

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^{- 6}

1.82 \times 10^{- 7}

1.82 \times 10^{- 8}

1.82 \times 10^{- 9}

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detailed solution

Correct option is D

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