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Electrolysis and Electrolytes

The limiting ...

Question

The limiting molar conductivity for KCl, NaBr and NaCl are 100, 120, 130 S cm^-2mol^-1 respectively. Calculate the limiting molar conductivity for KBr?

110 S cm^-2mol^-1

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210 S cm^-2mol^-1

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350 S cm^-2mol^-1

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90 S cm^-2mol^-1

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Solution

The correct option is D
90 S cm^-2mol^-1

We know from Kohlrausch law that for

E⁰(KCl) = λ⁰_Cl- + λ⁰_K+ = 100 scm^-2mol^-1

E⁰(NaBr) = λ⁰_Br- + λ⁰_Na+ = 120 scm^-2mol^-1

E⁰(NaCl) = λ⁰_Na+ + λ⁰_Cl- = 130 scm^-2mol^-1

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

100 S cm^-2mol^-1 - 130 S cm^-2mol^-1 + 120 S cm^-2mol^-1 = 90 S cm^-2mol^-1

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K C l

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S c m^{2} m o 1^{- 1}

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4 \times 10^{2} S c m^{2} m o l^{- 1}

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y

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N H_{4} O H

λ_{m}^{\infty} for B a (O H)_{2} = 457.6 o h m^{- 1} c m^{2} m o l^{- 1} .

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