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Q.

The conductivity of 0.001M acetic acid is $5 \times 10^{- 5} S C M^{- 1}$ and $μ_{0}$ is $390.5 S C M^{2} M o l^{- 1}$ then the calculated value of dissociation constant of acetic acid would be,

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a

$8 \times 10^{- 6}$

b

$18 \times 10^{- 5}$

c

$18 \times 10^{- 6}$

d

$8 \times 10^{- 4}$

answer is C.

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Detailed Solution

$^_{M} = \frac{1000 \times k}{M},$

$^_{M} = \frac{10^{3} \times 5 \times 10^{- 5}}{10^{- 3}}$

$α = \frac{^_{0}}{^_{M}}$

$k = c α^{2}$ = $18 \times 10^{- 6}$

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