G Mol To Mol L
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I know I should dissever $\pu{g L-one}$ by the molar mass of the substance, but I don't seem to find the specific answer on Google. So but to be sure:
If I take $10^{-5}~\pu{g L-1}~\ce{Cu^two+}$ solution, do I have $one.57 \times 10^{-7}~\pu{mol L-1}$?
I think I'one thousand missing something.
Sir Arthur7
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asked May six, 2015 at 14:07
Ndrina LimaniNdrina Limani
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two Answers 2
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Yes. $$n = \frac{thousand}{M}~~~~~~~~~~n = cV$$ So $$c = \frac{1000}{VM}$$
Given $\frac{thou}{V}$ yous can piece of work out $c$ (in $\mathrm{mol~dm^{-3}}$).
answered May 6, 2015 at 14:15
bonbon
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Grams of $\ce{Cu^{+2}}$ is sick divers. The $\ce{Cu^{+2}}$ did non get into the solution by itself. It was $\ce{CuCl2,CuSO4}$ or some other common salt.
This is probably the reason for the discrepancy.
answered May 7, 2015 at 10:28
Burak UlgutBurak Ulgut
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$\begingroup$ What discrepancy? The OP's question seems perfectly clear to me. $\endgroup$
May 7, 2015 at xv:55
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$\begingroup$ The OP is mentioning a different answer compared to "the specific answer on google". That is what I meant by
discrepancy$\endgroup$May 8, 2015 at 5:55
G Mol To Mol L,
Source: https://chemistry.stackexchange.com/questions/30945/how-to-convert-g-l-to-mol-l
Posted by: cobbalkinst.blogspot.com
$\begingroup$ Dimensional assay is helpful. g/50 divided by thou/mol gives mol/L. $\endgroup$
Jun 12, 2020 at 18:40