Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 decade ago

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

NiCl2 from the salt NiCl * 6H2O

sodium carbonate from the pure solid

please explain

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    (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

    --

    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

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