Answer:
0.0428 M
Step-by-step explanation:
Because we're asked to calculate the molarity of nickel(II) cation, we need to determine all sources for that species, in this case, all Ni⁺² comes from the nickel(II) bromide solid (NiBr₂).
We use the molecular weight of NiBr₂ to calculate the moles of Ni:
1.87 g NiBr₂ ÷ 218.49g/mol * (1molNi⁺²/1molNiBr₂) = 8.55x10⁻³ mol Ni⁺²
Then we divide the moles by the volume in order to calculate the concentration:
8.55x10⁻³ mol Ni⁺² / 0.200 L = 0.0428 M