Question
Write a mote on molar conductivity $\left(\wedge_{\text {m }}\right)$ of solution and write down it's formula.
✓
Answer
→ The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to charge and size of the ions in which they dissociate, the concentration of ions or ease with which the ions move under a potential gradient.
→ It, therefore becomes necessary to define physically more meaningful quantity called molar conductivity denoted by the symbol ^m (Greek, lambda)
Molar conductivity $=\wedge_{ m }=\frac{\kappa}{ c }$
→ In the above equation, if k is expressed in $S$ $m ^{-1}$ and the concentration, c in mol $m ^{-3}$ then the units of $\wedge_{ m }$ are in $S\ m ^2\ mol^{-1}$. It may be noted that :
$1 \ mol m ^{-3}=1000\left(L / m ^3\right) \times$ molarity $( mol / L )$, and hence $
→ \wedge_{ m }\left( S\ m ^2mol^{-1}\right)=\frac{\kappa Sm ^{-1}}{1000 Lm ^{-3} \times \text { molarity } mol L ^{-1}}
$
→ If we use $S cm ^{-1}$ as the units for k and mol $cm ^3$, the units of concentration, then the units for $\wedge_{ m }$ are $S cm ^2 mol^{-1}$.It can be calculated by using the equation :
$
\begin{array}{l}
\wedge_{ m }\left( S cm ^2\ mol^{-1}\right)=\frac{\kappa S cm ^{-1} \times 1000 cm^3 L}{ molarity mol L } \\
1\ S\ m ^2\ mol^{-1}=10^4 \ S \ cm ^2 \ mol^{-1} \text { or } \\
1 \ S\ cm ^2 \ mol^{-1}=10^{-4} \ S\ m ^2\ mol^{-1}
\end{array}
$
→ It, therefore becomes necessary to define physically more meaningful quantity called molar conductivity denoted by the symbol ^m (Greek, lambda)
Molar conductivity $=\wedge_{ m }=\frac{\kappa}{ c }$
→ In the above equation, if k is expressed in $S$ $m ^{-1}$ and the concentration, c in mol $m ^{-3}$ then the units of $\wedge_{ m }$ are in $S\ m ^2\ mol^{-1}$. It may be noted that :
$1 \ mol m ^{-3}=1000\left(L / m ^3\right) \times$ molarity $( mol / L )$, and hence $
→ \wedge_{ m }\left( S\ m ^2mol^{-1}\right)=\frac{\kappa Sm ^{-1}}{1000 Lm ^{-3} \times \text { molarity } mol L ^{-1}}
$
→ If we use $S cm ^{-1}$ as the units for k and mol $cm ^3$, the units of concentration, then the units for $\wedge_{ m }$ are $S cm ^2 mol^{-1}$.It can be calculated by using the equation :
$
\begin{array}{l}
\wedge_{ m }\left( S cm ^2\ mol^{-1}\right)=\frac{\kappa S cm ^{-1} \times 1000 cm^3 L}{ molarity mol L } \\
1\ S\ m ^2\ mol^{-1}=10^4 \ S \ cm ^2 \ mol^{-1} \text { or } \\
1 \ S\ cm ^2 \ mol^{-1}=10^{-4} \ S\ m ^2\ mol^{-1}
\end{array}
$
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