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

A
$1.82 \times 10^{- 6}$
B
$1.82 \times 10^{- 7}$
C
$1.82 \times 10^{- 8}$
D
$1.82 \times 10^{- 9}$

Solution:

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

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

Solution:

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