Rajasthan BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
Write a mote on molar conductivity $\left(\wedge_{\text {m }}\right)$ of solution and write down it's formula.
✓
Answer
$\rightarrow$ 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.
$\rightarrow$ It, therefore becomes necessary to define physically more meaningful quantity called molar conductivity denoted by the symbol$^m\ ($Greek, lambda$)$
Molar conductivity $=\wedge_{ m }=\frac{\kappa}{ c }$
$\rightarrow$ 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
$\rightarrow \wedge_{ m }\left( S\ m ^2mol^{-1}\right)=\frac{\kappa Sm ^{-1}}{1000 Lm ^{-3} \times \text { molarity } mol L ^{-1}}$
$\rightarrow$ 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 :
$\wedge_{ m }\left( S \ cm ^2\ mol^{-1}\right)=\frac{\kappa S \ cm ^{-1} \times 1000 \ cm^3 L}{ molarity mol L }$
$1\ S\ m ^2\ mol^{-1}=10^4 \ S \ cm ^2 \ mol^{-1} \text { or }$
$1 \ S\ cm ^2 \ mol^{-1}=10^{-4} \ S\ m ^2\ mol^{-1}$
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