Question

1. What mass (in g) of nickel (II) hexahydrate will you need for this solution? (the molar mass of nickel II hexahydrate is 290.79 g/mol)

2. If you had a 0.500 M stock solution, how many mL will you need from the stock to prepare 25.0 mL of 0.125 M?

Please provide a detailed explanation with your answer.

Answer 1

1. This question need more information such as concentration and volume of solution.

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2.

Given data

Molarity of given stock solution = 0.5 M

Molarity of solution to be prepared = 0.125 M

Volume of of solution to be prepared = 25.0 mL

                                                   =0.025 L

Moles of Ni(ii)hexahydrate required to prepare 25 mL of 0.125 M solution

                                                  = molarity * volume in liters

                                                  = 0.125 M * 0.025 L

                                                  = 0.003125 mol

       Volume of 0.5 M Ni(ii)hexahydrate solution required to get 0.003125 moles

                                      

Volume of 0.5 M Ni(ii)hexahydrate required to prepare 25 mL of 0.125 M solution = 6.25 mL