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Step-by-Step Solution
Step 1: Identify the Given Data
• Equivalent conductance of $Ba^{2+}$ at infinite dilution, $ \lambda_{\infty}(Ba^{2+}) = 127\ \mathrm{\Omega^{-1}cm^2\,eq^{-1}}$
• Equivalent conductance of $Cl^{-}$ at infinite dilution, $ \lambda_{\infty}(Cl^{-}) = 76\ \mathrm{\Omega^{-1}cm^2\,eq^{-1}}$
Step 2: Recall the Relevant Concept (Kohlrausch’s Law)
For an electrolyte $BaCl_{2}$ that dissociates as
$ BaCl_{2} \rightarrow Ba^{2+} + 2\,Cl^{-}, $
the molar conductivity at infinite dilution is the sum of the ionic conductivities multiplied by their respective coefficients:
$ \Lambda_m^{\infty}(BaCl_{2}) = \lambda_{\infty}(Ba^{2+}) + 2\,\lambda_{\infty}(Cl^{-}) \,. $
However, the question asks for the equivalent conductance, $ \lambda_{\infty}(BaCl_{2}) $. Since 1 mole of $BaCl_{2}$ corresponds to 2 equivalents (because $Ba^{2+}$ has a charge of +2), the relationship between molar conductance and equivalent conductance is:
$ \lambda_{\infty}(BaCl_{2}) = \frac{\Lambda_m^{\infty}(BaCl_{2})}{2} \,. $
Step 3: Substitute the Values
First, calculate the molar conductivity at infinite dilution:
$ \Lambda_m^{\infty}(BaCl_{2}) = \lambda_{\infty}(Ba^{2+}) + 2\,\lambda_{\infty}(Cl^{-}) \\[6pt]
= 127 + 2 \times 76 \\[6pt]
= 127 + 152 \\[6pt]
= 279\ \mathrm{\Omega^{-1}cm^2\,mol^{-1}};
$
Now, convert it to equivalent conductance:
$ \lambda_{\infty}(BaCl_{2}) = \frac{279}{2} = 139.5\ \mathrm{\Omega^{-1}cm^2\,eq^{-1}}.
$
Step 4: State the Final Answer
The equivalent conductance of $BaCl_{2}$ at infinite dilution is 139.5 $\mathrm{\Omega^{-1}cm^2\,eq^{-1}}$.
