Molar conductance of decimolar electrolyte is $' X' {mhocm}^{2} {mole}^{- 1}$ . Specific conductance would be _____ ${mhocm}^{- 1}$

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$\frac{10^{- 2}}{X}$

$10^{- 4} X$

$10^{- 2} X$

$10^{- 3} X$

answer is D.

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Detailed Solution

$\begin{array}{l} λ_{m} ({mhocm}^{2} {mole}^{- 1}) = K ({mhocm}^{- 1}) \times \frac{1000}{M} \\ K ({mhocm}^{- 1}) = \frac{X \times 0 .1}{1000} = 10^{- 4} X \end{array}$

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