First slide

Applications of Kohlraush law

Question

If the molar conductance values of Ca²⁺ and Cl^– at infinite dilution are respectively 118.88 × 10^-4 m² mho mol^-1 and 77.33 × 10^-4 m² mho mol^-1, then that of CaCl₂ is (in m² mho mol^-1)

Easy

Solution

$CaC{l_2} \rightleftharpoons C{a^{ + 2}} + 2C{l^ - }$

$\lambda _{CaC{l_2}}^m = \lambda _{C{a^{ + 2}}}^m + 2\lambda _{C{l^ - }}^m$

$\lambda _{CaC{l_2}}^m = \left( {118.88 \times {{10}^{ - 4}}} \right) + 2\left( {77.33 \times {{10}^{ - 4}}} \right) = 273.54 \times {10^{ - 4}}m{\hom ^2}mol{e^{ - 1}}$

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