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(B) The molar conductivity at infinite dilution $(\Lambda _m^\infty)$ is related to the equivalent conductivity at infinite dilution $(\lambda _{eq}^\

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$\Lambda _{m_{CaF_2}}^\infty = 2\left( \lambda _{eq_{Ca^{2+}}}^\infty + \lambda _{eq_{F^{-}}}^\infty \right)$

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(B) The molar conductivity at infinite dilution $(\Lambda _m^\infty)$ is related to the equivalent conductivity at infinite dilution $(\lambda _{eq}^\infty)$ by the formula: $\Lambda _m^\infty = n 	imes \lambda _{eq}^\infty$,where $n$ is the valency factor (total positive or negative charge per formula unit). For $CaF_2$,the dissociation is $CaF_2 ightarrow Ca^{2+} + 2F^-$. The valency factor $n$ for $CaF_2$ is $2$. Thus,$\Lambda _{m_{CaF_2}}^\infty = 2 	imes \lambda _{eq_{CaF_2}}^\infty$. According to Kohlrausch's law,$\lambda _{eq_{CaF_2}}^\infty = \lambda _{eq_{Ca^{2+}}}^\infty + \lambda _{eq_{F^{-}}}^\infty$. Substituting this,we get $\Lambda _{m_{CaF_2}}^\infty = 2(\lambda _{eq_{Ca^{2+}}}^\infty + \lambda _{eq_{F^{-}}}^\infty)$.

For a saturated solution of $CaF_2$,choose the correct relationship-

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