Question #58011
You need to make an aqueous solution of 0.123 M nickel(II) fluoride for an experiment in lab, using a 500 mL volumetric flask. How much solid nickel(II) fluoride should you add?
Expert's answer
The amount of substance of NiF2 is:
n(NiF2)=V*CM(NiF2) = 0.5 L*0.123 M= 0.0615 mol
The mass of NiF2 is:
m(NiF2) = M(NiF2)* n(NiF2) = 97 g/mol* 0.0615 mol = 5.9655 g
Answer:
We need to add 5.9655 g of nickel(II) fluoride
n(NiF2)=V*CM(NiF2) = 0.5 L*0.123 M= 0.0615 mol
The mass of NiF2 is:
m(NiF2) = M(NiF2)* n(NiF2) = 97 g/mol* 0.0615 mol = 5.9655 g
Answer:
We need to add 5.9655 g of nickel(II) fluoride
