The conductivity of water at 298 K is 5.54×10−8Scm−1. If λH+∞=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−$$1=1.82$$×$$10$$−$$9$$

Q.

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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By Expert Faculty of Sri Chaitanya

$1.82 \times 10^{- 6}$

$1.82 \times 10^{- 7}$

$1.82 \times 10^{- 8}$

$1.82 \times 10^{- 9}$

answer is D.

(Detailed Solution Below)

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Detailed Solution

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

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