Question

If the molar conductance values of $C{a}^{2+}$ and $C{l}^{-}$ at infinite dilution are respectively $118.88\times {10}^{-4}{m}^{2}mhomo{l}^{-1}$ and $77.33\times {10}^{-4}{m}^{2}mhomh{o}^{-1}$ then that of $CaC{l}_2$ is:

• A. $118.88\times {10}^{-4}$
• B. $154.66\times {10}^{-4}$
• C. $273.54\times {10}^{-4}$
• D. $196.21\times {10}^{-4}$

Answer 1

The correct option is C $273.54\times {10}^{-4}$

${\wedge }_m^{\circ }CaC{l}_2={\lambda }^{\circ }C{a}^{2+}+2{\lambda }^{\circ }C{l}^{-}$
$=(118.88\times \times {10}^{-4})+2(77.33\times {10}^{-4}$
$=273.54\times {10}^{-4}{m}^{2}mhomo{l}^{-1}$.