Question

At infinite dilution equivalent conductances of $Ba^{2+}$ & $Cl^–$ ions are $127$ & $76$ $ohm^{–1}cm^{–1}eq^{–1}$ respectively. Equivalent conductance of $BaCl_2$ at infinite dilutions is:

• A. 139.5
• B. 101.5
• C. 203
• D. 279

Answer 1

Answer: (A)

139.5

Answer 2

The chemical equation is,
$BaCl_2 → Ba^{2+} + 2Cl^-$
So,$\ Λ_{BaCl_2}^0​=Λ_{Ba^{2+}}^0​+2Λ_{Cl^−}^0​$
Given that,
$Λ_{Ba^{2+}}^0​=127\ Ω^{−1}cm^2eq^{−1}$
$Λ_{Cl^−}^0​=76\ Ω^{−1} cm^{2}eq^{−1}$
$Λ_{BaCl_2}^0​$ ​$= 127+2\times 76 = 127+152 = 279\ Ω^{−1}cm^2eq^{−1}$
The equivalent conductance of $BaCl_2$ at infinite dilution = $\frac {279}{2}$ = $139.5\ Ω^{−1}cm^2eq^{−1}$

So, the correct option is (A): 139.5.