[xc630]

Hi

The question asks how would you prepare 1.00L of a 0.50 M solution of each of the following:

Some they list in the format such as HCl from concentrated 12 M reagent.  I just use M1V1 =M2V2 for these.
However how would I do: NiCl2 from the salt NiCl2 x 6H2O.  The "x" is really a dot suspended in between the two.  How ould I find the molarity just from that?

Another lists: sodium carbonate from the pure solid.  Do I just find the moles from the molecular weight in this case?

thanks

[Dan]

However how would I do: NiCl2 from the salt NiCl2 x 6H2O.  The "x" is really a dot suspended in between the two.  How ould I find the molarity just from that?

Another lists: sodium carbonate from the pure solid.  Do I just find the moles from the molecular weight in this case?

thanks

about the sodium carbonate, yes that's right.
Use the same principle for the NiCl₂.6H₂O

[xc630]

for the NiCl2 do I just ignore the 6H2O?

[Yggdrasil]

You need to take into account the 6H₂O when calculating the molecular weight of the NiCl₂·6H₂O.

(In case you didn't know, the dot just means that your nickle chloride salt is hydrated.  For every unit of NiCl₂, there are six molecules of water bound)