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Question

(A) According to Kohlrausch's law of independent migration of ions: $\wedge^0_{NaBr} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$ $\wedge^0_{NaCl} = \lambda

Vedclass · Accepted Answer

$128 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

Vedclass · Answer

(A) According to Kohlrausch's law of independent migration of ions: $\wedge^0_{NaBr} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$ $\wedge^0_{NaCl} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ $\wedge^0_{KBr} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ $\wedge^0_{KCl} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ To obtain $\wedge^0_{NaBr}$,we perform the operation: $\wedge^0_{NaBr} = \wedge^0_{NaCl} + \wedge^0_{KBr} - \wedge^0_{KCl}$ $\wedge^0_{NaBr} = 126 + 152 - 150 = 128 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

List-$I$ (Symbol of electrical property)	List-$II$ (Units)
$A.$ $\wedge_m$	$I.$ $S\,cm^2\,mol^{-1}$
$B.$ $G$	$II.$ $S$
$C.$ $\kappa$	$III.$ $S\,cm^{-1}$
$D.$ $G^*$	$IV.$ $cm^{-1}$

Calculate molar conductivity at infinite dilution for $NaBr$ if molar conductivity at infinite dilution for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

Similar Questions

What is the molar conductivity of $0.005 \ M$ $NaI$ solution if its conductivity is $6.065 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$?

Match the following:
List-$I$ (Symbol of electrical property) List-$II$ (Units)
$A.$ $\wedge_m$ $I.$ $S\,cm^2\,mol^{-1}$
$B.$ $G$ $II.$ $S$
$C.$ $\kappa$ $III.$ $S\,cm^{-1}$
$D.$ $G^*$ $IV.$ $cm^{-1}$

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