What is the molar conductivity at infinite di | vedclass

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Question

(C) According to Kohlrausch's Law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the mol

Vedclass · Accepted Answer

$261.0 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$

Vedclass · Answer

(C) According to Kohlrausch's Law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions.$\wedge_{m}^{0}(CaCl_{2}) = \lambda_{Ca^{2+}}^{0} + 2 \lambda_{Cl^{-}}^{0}$Given,$\lambda_{Ca^{2+}}^{0} = 119 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ and $\lambda_{Cl^{-}}^{0} = 71 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.Substituting the values:$\wedge_{m}^{0}(CaCl_{2}) = 119 + 2(71)$$\wedge_{m}^{0}(CaCl_{2}) = 119 + 142 = 261.0 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$.

List-$I$ (Parameter)	List-$II$ (Unit)
$a$. Cell constant	$i$. $S\, cm^{2}\, mol^{-1}$
$b$. Molar conductivity	$ii$. Dimensionless
$c$. Conductivity	$iii$. $m^{-1}$
$d$. Degree of dissociation of electrolyte	$iv$. $\Omega^{-1}\, m^{-1}$

What is the molar conductivity at infinite dilution of $CaCl_{2}$,if the molar conductivity of $Ca^{2+}$ ion and $Cl^{-}$ ion at infinite dilution is $119$ and $71 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ respectively?

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