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Could you explain what is unit of molar conductivity along with derivation

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Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution. It's unit in S.I. is: $$\ce{S.m^2.mol^{-1}}$$

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  • $\begingroup$ What is "S" over here? $\endgroup$ – Vidya Sagar V Jan 4 '15 at 6:02
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    $\begingroup$ S is Simens. It equals $Ohm^{-1}$ $\endgroup$ – Yomen Atassi Jan 4 '15 at 6:15
  • $\begingroup$ Is simens an S.I unit $\endgroup$ – Vidya Sagar V Jan 4 '15 at 6:16
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    $\begingroup$ Yes, as $Ohm$ is an derived S.I. unit. $\endgroup$ – Yomen Atassi Jan 4 '15 at 6:18
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unit of conductivity of an electrolyte solution =

(unit of length)/(unit of area) * (unit of Conductance)

= (m/m2). S = S . m-1

unit of molar conductivity = (unit of conductivity of an electrolyte solution ) / (unit of molar concentration of the electrolyte) = ( S m-1) / (mol m-3) = S . m . 2 mol-1

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  • $\begingroup$ This does not look like a full answer yet. Also, please check out what you can find on MathJax in the help center and this meta-post and the links therein to learn how to format your post with MathJax; it would improve the look of the answer =) $\endgroup$ – Jan Jan 5 '17 at 9:32
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It does look odd that the numerator is expressed in units of conductance times units of area. Conductivity in Siemens per meter divided by concentration in, say, moles per liter should give you something in units of $S \cdot l \cdot mol^{-1} \cdot m^{-1}$ . But since a liter is simply a volume measurement equal to 1000 $cm^3$ or 0.001 $m^3$, the volume/distance can be cancelled out to just area. So you end up with that $S \cdot m^2 \cdot mol^{-1}$.

Sometimes you see tables that contain molar conductivity values in units of $ 10^{-4} S \cdot m^2 \cdot mol^{-1}$. This makes it handy because the $10^{-4}$ factor gives you the same value as when expressed in units of $ \mu S/cm $ per $ mmol/l $.

E.g, the specific conductance for NaCl is:

$ 0.0126 \space S \cdot m^2 \cdot mol^{-1}$

$ 126 \times 10^{-4} \space S \cdot m^2 \cdot mol^{-1}$

$126$ $(\mu S / cm) \space \over \space (mmol / l) $

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