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Enthalpy of Neutralisation

The specific ...

Question

The specific conductance (K) of an electrolyte of 0.1 N concentration is related to equivalent conductance ( $Λ$ ) by the following

Λ = K

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Λ = 10 K

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Λ = 100 K

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Λ = 10000 K

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Solution

The correct option is D $Λ = 10000 K$
$Λ = \frac{1000 \times K}{N} Λ = \frac{1000 \times K}{0.1} = 10000 K$

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(K)

0.1 N

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According to Kohlrausch law, the limiting value of molar conductivity of an electrolyte A₂B is:

\frac{Λ_{v}^{2}}{Λ_{\infty} (Λ_{\infty} - Λ_{v})} v s . \frac{1}{c}

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The $Λ_{e q}^{\circ}$ for potash alum $(K_{2} S O_{4} . A l_{2} (S O_{4})_{3} . 24 H_{2} O)$ is :

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C

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