The equivalent conductance of C6H5COONa- HCl and NaCl at infinite dilution are 82-4- 426-2 and 126-5 - -1cm2- Calculate the equivalent conductance - -1cm2 of benzoic acid at infinite dilution-

From Kohlrauch's law, we get nbsp; λ ^∘ _C_6H_5COOH= λ ^∘ _C_6H_5COONa + λ ^∘ _HCl λ ^∘ _NaCl =82.4+ 426.2 126.5 = 382.1 Ω ^ 1cm^2 ...

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Standard VII

Chemistry

Ionic Conductance

The equivalen...

Question

The equivalent conductance of $C_{6} H_{5} C O O N a, H C l$ and $N a C l$ at infinite dilution are $82.4, 426.2$ and $126.5 Ω^{- 1} {cm}^{2}$ . Calculate the equivalent conductance $(Ω^{- 1} {cm}^{2})$ of benzoic acid at infinite dilution.

217.3 Ω^{- 1} {cm}^{2}

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272.1 Ω^{- 1} {cm}^{2}

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191.05 Ω^{- 1} {cm}^{2}

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382.1 Ω^{- 1} {cm}^{2}

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Solution

The correct option is D $382.1 Ω^{- 1} {cm}^{2}$
From Kohlrauch's law, we get
$λ_{C_{6} H_{5} C O O H}^{\circ}$ $= λ_{C_{6} H_{5} C O O N a}^{\circ} + λ_{H C l}^{\circ} - λ_{N a C l}^{\circ} = 82.4 + 426.2 - 126.5 = 382.1 Ω^{- 1} {cm}^{2}$

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The equivalent conductance of C6H5COONa, HCl and NaCl at infinite dilution are 82.4, 426.2 and 126.5 Ω−1cm2. Calculate the equivalent conductance (Ω−1cm2) of benzoic acid at infinite dilution.

The correct option is D 382.1 Ω−1cm2From Kohlrauch's law, we get λ∘C6H5COOH=λ∘C6H5COONa+λ∘HCl−λ∘NaCl=82.4+426.2−126.5=382.1 Ω−1cm2

The equivalent conductance of $C_{6} H_{5} C O O N a, H C l$ and $N a C l$ at infinite dilution are $82.4, 426.2$ and $126.5 Ω^{- 1} {cm}^{2}$ . Calculate the equivalent conductance $(Ω^{- 1} {cm}^{2})$ of benzoic acid at infinite dilution.

The correct option is D $382.1 Ω^{- 1} {cm}^{2}$
From Kohlrauch's law, we get
$λ_{C_{6} H_{5} C O O H}^{\circ}$ $= λ_{C_{6} H_{5} C O O N a}^{\circ} + λ_{H C l}^{\circ} - λ_{N a C l}^{\circ} = 82.4 + 426.2 - 126.5 = 382.1 Ω^{- 1} {cm}^{2}$