How many moles of ions are in 285 ml of 0.0150 m mgcl2?

MgCl₂) = Mg²⁺ + 2Cl⁻

V (MgCl₂) = 285cm³=0,285dm³

c (MgCl₂) = 0,015 mol/dm³

n (MgCl₂) = c·V = 0,015 mol/dm³ · 0,285dm³ = 0,0042 mol

n (Mg²⁺) = n (MgCl₂) = 0,0042 mol

n (Cl⁻) = 2n (MgCl₂) = 0,0084 mol