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# Calculate $\wedge^{\large\circ}_m$ for KCl solution if molar ionic conductances at infinite dilution are $7.352\times 10^{-3}Sm^2mol^{-1}$ for $K^+$ and $7.634\times 10^{-3}Sm^2mol^{-1}$ for $Cl^{-}$.The $\wedge^{\large\circ}_m$ is

$\begin{array}{1 1}(A)\;1.4986\times 10^{-2}Sm^2mol^{-1}&(B)\;1.2974\times 10^{-1}Sm^2mol^{-1}\\(C)\;6.43\times 10^{-3}Sm^2mol^{-1}&(D)\;9.24\times 10^{-2}Sm^2mol^{-1}\end{array}$

Can you answer this question?

$\wedge^{\large\circ}_m(KCl)=\wedge^{\large\circ}_{K^+}(aq)+\wedge^{\large\circ}_{Cl^-}(aq)$
$\qquad\;\;\quad=7.352\times 10^{-3}+7.634\times 10^{-3}$
$\qquad\;\;\quad=1.4986\times 10^{-2}Sm^2mol^{-1}$
Hence (A) is the correct answer.
answered Dec 10, 2013