A) \[118.88\times {{10}^{-4}}\]
B) \[154.66\times {{10}^{-4}}\]
C) \[273.54\times {{10}^{-4}}\]
D) \[196.21\times {{10}^{-4}}\]
Correct Answer: C
Solution :
From Kohlrauschs law: \[\wedge _{m}^{\infty }={{v}_{+}}\lambda _{+}^{\infty }+{{v}_{-}}\lambda _{-}^{\infty }\] For \[CaC{{l}_{2}}\] \[\wedge _{m}^{\infty }(CaC{{l}_{1}})=\lambda _{C{{a}^{2+}}}^{\infty }+2\lambda _{C{{l}^{-}}}^{\infty }\] \[=118.88\times {{10}^{-4}}+2\times 77.33\times {{10}^{-4}}\] \[=118.88\times {{10}^{-4}}+154.66\times {{10}^{-4}}\] \[=273.54\times {{10}^{-4}}{{m}^{2}}\,mho\,mo{{l}^{-1}}\]
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