The limiting molar conductivity for AB - BC and CD are in the ratio of 1- 3- 5 S cm -2 mol -1 respectively- Calculate the -0 for AD-A- 100 S cm -2 mol -1B- 450 S cm -2 mol -1C- 300 S cm -2 mol -1D- 900 S cm -2 mol -1

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Linear Dependence and Independence of Vectors

The limiting ...

Question

The limiting molar conductivity for AB , BC and CD are in the ratio of 1:3:5 S cm^-2mol^-1 respectively. Calculate the Λ⁰ for AD.

100 S cm^-2mol^-1

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450 S cm^-2mol^-1

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300 S cm^-2mol^-1

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900 S cm^-2mol^-1

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Solution

The correct option is C
300 S cm^-2mol^-1

We know from Kohlrausch law that for

Λ⁰ (AB) = λ⁰_B^- + λ⁰_A⁺ = 100S cm^-2mol^-1 ….(i)

Λ⁰ ( BC) = λ⁰_C^- + λ⁰_B⁺ = 300S cm^-2mol^-1 …………. (ii)

Λ⁰ ( CD) = λ⁰_C⁺ + λ⁰_D^- = 500 S cm^-2mol^-1 ………….(iii)

Subtracting ( iii) from ( i) and add (ii) to it

100 S cm^-2mol^-1 + 500 S cm^-2mol^-1 - 300 S cm^-2mol = 300 S cm^-2mol

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The limiting molar conductivity for AB , BC and CD are in the ratio of 1:3:5 S cm-2 mol-1 respectively. Calculate the Λ0 for AD.

The limiting molar conductivity for AB , BC and CD are in the ratio of 1:3:5 S cm^-2mol^-1 respectively. Calculate the Λ⁰ for AD.