Question

The ionic conductivity of $B{a}^{2+}$ and $C{l}^{-}$ at inifnite dilution are 127 and 76 ${\text{ohm}}^{-1}{\text{cm}}^2{\text{mol}}^{-1}$ respectively. The equivalent conductivity of $BaC{l}_2$ at inifinity dilution (in ${\text{ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}$) would be:

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

Answer 1

The correct option is D $139.5$

Given,
molar ionic conductivity of $B{a}^{2+}$
${\lambda }_{\left[B{a}^{2+}\right]}^{\infty }=127{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}$
molar ionic conductivity of $C{l}^{-}$
${\lambda }_{\left[C{l}^{-}\right]}^{\infty }=76{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}$
Therefore
${\Lambda }_{\left[BaC{l}_2\right]}^{\infty }={\lambda }_{\left[B{a}^{2+}\right]}^{\infty }+2\times {\lambda }_{\left[C{l}^{-}\right]}^{\infty }$
$=(127+2\times 76){\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}$
$=279{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}$
Equivalent condutance $=$ $\frac{\text{molar conductance}}{\text{n-factor}}=\frac{279}{2}=139.5{\Omega }^{-1}{\text{cm}}^2{\text{eq}}^{-1}$