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At infinite dilution equivalent conductances of $Ba^{2+}$ and $Cl^-$ ions are 127 and 76 ohm$^{-1}$ cm$^{2}$ eq$^{-1}$ respectively. Equivalent conductance of $BaCl_2$ at infinite dilution is :
A
139.5
B
101.5
C
203
D
279
Detailed Solution
By Kohlrausch's law, the limiting equivalent conductance of an electrolyte is the sum of the limiting equivalent conductances of its ions.
The value 127 for barium corresponds to the whole $Ba^{2+}$ ion; per equivalent it is $\dfrac{1}{2} \times 127 = 63.5$.
$\Lambda^\circ_{eq}(BaCl_2) = \lambda^\circ\left(\dfrac{1}{2}Ba^{2+}\right) + \lambda^\circ(Cl^-) = 63.5 + 76$
$= 139.5$ ohm$^{-1}$ cm$^2$ eq$^{-1}$
The value 127 for barium corresponds to the whole $Ba^{2+}$ ion; per equivalent it is $\dfrac{1}{2} \times 127 = 63.5$.
$\Lambda^\circ_{eq}(BaCl_2) = \lambda^\circ\left(\dfrac{1}{2}Ba^{2+}\right) + \lambda^\circ(Cl^-) = 63.5 + 76$
$= 139.5$ ohm$^{-1}$ cm$^2$ eq$^{-1}$