Question
Explain kohlrausch Law with example.
✓
Answer
\rightarrow "The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte."
\rightarrow Thus, if $\lambda_{ Na ^{+}}^0$ and $\lambda_{ Cl ^{-}}^0$ are limiting molar conductivity of the sodium and chloride ions respectively. then the limiting molar conductivity for sodium chloride is given by the equation :
$\wedge_{ m ( NaCl )}^0=\lambda_{ Na ^{+}}^0+\lambda_{ Cl ^{-}}^0$
$\rightarrow$ In general, if an electrolyte on dissociation gives $v_{+}$cations and $v_{-}$anions then its limiting molar conductivity is given by :
$E _{ m }^0= v _{+} \lambda_{+}^0+ v _{-} \lambda^0$
$\rightarrow$ Here, $\lambda_{+}^0$ and $\lambda_{-}^0$ are the limiting molar conductivities of the cation and anion respectively.
$\rightarrow \text { If } \lambda_{ m ^{+}}^0 \text { for } Na ^{+}=50.1 \ S\ cm ^2\ mol^{-1}$
$\lambda_{ m ^{-}}^0 \text { for } Cl ^{-}=76.3 \ S \ cm ^2 \ mol^{-1}$
$\lambda_{ m ( NaCl )}=\lambda_{ m \left( Na ^{+}\right)}^0+\lambda_{ m \left( Cl ^{-}\right)}^0$
$=50.1+76.3$
$=126.4 \ S \ \cdot \ cm ^2 \ mol^{-1}$
\rightarrow Thus, if $\lambda_{ Na ^{+}}^0$ and $\lambda_{ Cl ^{-}}^0$ are limiting molar conductivity of the sodium and chloride ions respectively. then the limiting molar conductivity for sodium chloride is given by the equation :
$\wedge_{ m ( NaCl )}^0=\lambda_{ Na ^{+}}^0+\lambda_{ Cl ^{-}}^0$
$\rightarrow$ In general, if an electrolyte on dissociation gives $v_{+}$cations and $v_{-}$anions then its limiting molar conductivity is given by :
$E _{ m }^0= v _{+} \lambda_{+}^0+ v _{-} \lambda^0$
$\rightarrow$ Here, $\lambda_{+}^0$ and $\lambda_{-}^0$ are the limiting molar conductivities of the cation and anion respectively.
$\rightarrow \text { If } \lambda_{ m ^{+}}^0 \text { for } Na ^{+}=50.1 \ S\ cm ^2\ mol^{-1}$
$\lambda_{ m ^{-}}^0 \text { for } Cl ^{-}=76.3 \ S \ cm ^2 \ mol^{-1}$
$\lambda_{ m ( NaCl )}=\lambda_{ m \left( Na ^{+}\right)}^0+\lambda_{ m \left( Cl ^{-}\right)}^0$
$=50.1+76.3$
$=126.4 \ S \ \cdot \ cm ^2 \ mol^{-1}$
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