If the molar conductance value of ${Ca}^{2 +}$ and ${Cl}^{-}$ at infinite dilution are respectively $118.88 imes 10^{- 4}\ m^{2}\ mho\ mol^{- 1}\$ and 77.33 $ imes 10^{- 4}\ m^{2}\ mho\ mol^{- 1}\$ then that of $CaCl_{2}$ is (in $m^{2}\ mho\ mol^{- 1}\$ )

Question

If the molar conductance value of ${Ca}^{2 +}$ and ${Cl}^{-}$ at infinite dilution are respectively $118.88 	imes 10^{- 4}\ m^{2}\ mho\ mol^{- 1}\$ and 77.33 $	imes 10^{- 4}\ m^{2}\ mho\ mol^{- 1}\$ then that of $CaCl_{2}$ is (in $m^{2}\ mho\ mol^{- 1}\$ )

Dr. Abubacker · Accepted Answer

(c)
From Kohlrausch’s law
$\Lambda_{m}^{\infty} = \ v_{+}\ \lambda_{+}^{\infty} + \ v_{-}\ \lambda_{-}^{\infty}$
For ${CaCl}_{2}$
$\Lambda_{m}^{\infty}({CaCl}_{2})\ \  = \ \Lambda_{Ca^{2 +}}^{\infty} + \ {2\Lambda}_{{Cl}^{-}}^{\infty}$
$= \ 118.88\  	imes \ 10^{- 4\ } + \ 2\  	imes \ 77.33\  	imes \ 10^{- 4}$
$= \ 118.88\  	imes \ 10^{- 4}\ \  + \ 154.66\  	imes \ 10^{- 4}$
$= \ 273.54\  	imes 10^{- 4}\ \ m^{2}\ mho\ {mol}^{- 1}$

Q. If the molar conductance value of ${Ca}^{2 +}$ and ${Cl}^{-}$ at infinite dilution are respectively $118.88 \times 10^{- 4}\ m^{2}\ mho\ mol^{- 1}\$ and 77.33 $\times 10^{- 4}\ m^{2}\ mho\ mol^{- 1}\$ then that of $CaCl_{2}$ is (in $m^{2}\ mho\ mol^{- 1}\$ )

Explanation

Reviewed by Dr. Abubacker

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