How would you prepare 1.00 L of a 0.50 M solution of each of the following?

Question

NiCl2 from the salt NiCl * 6H2O

sodium carbonate from the pure solid

please explain

Answer 1

(0.50 mol NiCl/L solution)*(1.0 L solution)*(237.69 g NiCl*6H2O/mole NiCl*6H2O)*(1 mole NiCl*6H2O/1 mole NiCl)= 118.84 g NiCl*6H2O.

          

So, weight out 118.84 grams of  NiCl2·6H2O , which I would dissolve into some water (enough that you can shake and dissolve) then dilute to 1.00 L

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Sodium carbonate has a  molar mass of 105.99 g/mol

0.50 moles* 105.99 g/ mol = 52.995 grams

Weight out 52.99 grams of Na2CO3 , & dissolve in some water (enough to shake it a bit) then dilute to 1.00 L