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I know I should divide $\mathrm{g~L^{-1}}$ by the molar mass of the substance, but I don't seem to find the specific answer on Google. So just to be sure:

If I have $10^{-5}~\mathrm{g~L^{-1}}~\ce{Cu^2+}$ solution, do I have $1.57 \times 10^{-7}~\mathrm{mol~L^{-1}}$?

I think I'm missing something.$%edit$

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Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$ So $$c = \frac{m}{VM}$$

Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).

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Grams of $\ce{Cu^{+2}}$ is ill defined. The $\ce{Cu^{+2}}$ did not get into the solution by itself. It was $\ce{CuCl2,CuSO4}$ or some other salt.

This is probably the reason for the discrepancy.

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  • $\begingroup$ What discrepancy? The OP's question seems perfectly clear to me. $\endgroup$ – bon May 7 '15 at 15:55
  • $\begingroup$ The OP is mentioning a different answer compared to "the specific answer on google". That is what I meant by discrepancy $\endgroup$ – Burak Ulgut May 8 '15 at 5:55

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