How many moles of ZnF2 are in 0.230 L of a 0.1125 M ZnF2 solution?

Answer 1

How many moles of ZnF2 are in 0.230 L of a 0.1125 M ZnF2 solution?

M = mol/V so mol = M x V = 0.1125M x 0.230 L = 0.02588 mol (0.0259 with the correct significant figures)