I know this might be a silly question but I am confused with different answers of internet. My question is:

Molar conductivity is given as:

Molar conductivity = $\frac{\kappa}{M}$.

S.I. unit of : $\kappa$ is $\pu{Sm}^{-1}$ &

$M$ is in $\mathrm{mol \, m^{-3}}$.

Thus, Molar conductivity in S.I. is $\mathrm{S \, m^2 \, mol^{-1}} \tag{1}$

When I substitute 1 m = 100 cm in (1). I got new unit as $[\mathrm{S \, cm^2 \, mol^{-1}}] \times10^4$. But in a book It was given as

$[\mathrm{S \, cm^2 \, mol^{-1}}] \times 10^3 \tag{2}$

where they have taken S in S/cm and molarity in mol/litre. But where I was wrong.

If you feel the question too confusing just explain me How I can convert that SI unit given in (1) into unit given in (2).

  • $\begingroup$ If M is molarity then yes it is taken in mol/litre and you have taken it mol/${m^3}$ which you can't do because it is defined that way. $\endgroup$ – Raghav Feb 1 '17 at 17:44
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    $\begingroup$ But SI unit of M is actually moles/m3. And hence for SI unit of molar conductivity it must include M in moles/m3 & k must be in S/m. Then only we get SI unit of molar conductivity as Sm^2/mol. $\endgroup$ – Avi Feb 2 '17 at 3:35

Refer to page 77 of Chemistry NCERT text book, class 12, part 1.

You were actually right about the $\mathrm{10^4\ S\ cm^2/mol =S\ m^2 /mol}$. Check if the $\mathrm{10^3}$ you are referring to has molarity included in its equation because molarity is given as moles per litre.

You can refer to the online NCERT book at its official website. Thank you, I hope this was useful and sorry if I got it wrong.

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    $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. Furthermore, please visit this page, this page and this ‎one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – Linear Christmas Feb 3 '17 at 18:59

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