Electrochemistry - Result Question 10
####11. The ionic conductance of $Ba^{2+}$ and $Cl^{-}$are respectively 127 and $76 ohm^{-1} cm^{2}$ at infinite dilution. The equivalent conductance (in ohm ${ }^{-1} cm^{2}$ ) of $BaCl_2$ at infinite dilution will be :
(a) 139.5
(b) 203
(c) 279
(d) 101.5
[2000]
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Solution:
- (b) The ionic conductance or equivalent conductance are given as:
$\lambda _{B a^{2+}}^{\infty}=127$ ohm $^{-1} cm^{2}$ equivalent ${ }^{-1}$
$\lambda _{Cl^{-}}^{\infty}=76$ ohm $^{-1} cm^{2}$ equivalent ${ }^{-1}$
The equivalent conductance of $BaCl_2$ at infinite dilution will be given by:
$ \begin{aligned} & \wedge_BaCl_2^{\infty}=\lambda _{Ba^{2}}^{\infty}+\lambda _{Cl^{-}}^{\infty}=127+76 \\ & =203 ohm^{-1} cm^{2} \text{ equivalent }{ }^{-1} \end{aligned} $
If equivalent conductance (conductivity) of cation and anion are $\lambda _{+e q}^{\infty}$ and $\lambda _{-e q}^{\infty}$ and equivalent conductance of electrolyte is $\lambda_eq^{\infty}$ in infinite dilution, then,
$ \wedge_eq^{\infty}=\lambda _{+eq}^{\infty}+\lambda _{-eq}^{\infty} $
At infinite dilution, if molar conductances of cation, anion and electrolyte are $\lambda _{+m}^{\infty}, \lambda _{-m}^{\infty}$ and $\lambda_m^{\infty}$ respectively for the electrolyte $A_x B_y$, then $A_x B_y \rightarrow x A^{y+}+y B^{x-}$
$\lambda _{\text{eq }}^{\infty}=\frac{1}{y} \lambda _{+}^{\infty} m+\frac{1}{x} \lambda _{-}^{\infty} m$ and
$\lambda_m^{\infty}=x \lambda _{+m}^{\infty}+y \lambda _{-}^{\infty} m$
