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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 $\mathrm{mol~dm^{-3}}$).
Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$So $$c = \frac{m}{VM}$$
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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.
2026-07-31
Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$ So $$c = \frac{m}{VM}$$
Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).
Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$So $$c = \frac{m}{VM}$$
Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).
More
VOTE
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.
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.
More
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