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.$%edit$


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

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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.