Question

Determine the molar concentration of the anion (negative ion), if 1.2g of magnesium chloride, MgCl2(s) is dissolved in enough water to make 2.0L of solution.

Answer 1

magnesium Chloride dissolves in water to form Mg²⁺ ions and Cl^- ions in a reversible process.

MgCl₂(s) $\iff$ Mg²⁺ (aq) + 2Cl^- (aq) in presence of water

mass of MgCl₂ = 1.2g and the volume of solution =2.0 L

$molarity = (Moles of solute)/ (litre volume of solution)$

calculating moles of magnesium Chloride

$mol = mass/molecular weight$

molecular weight of MgCl₂ = 24.0 +35.5 +35.5 = 95g

$moles = 1.2 / 95 = 0.01263$

mole ratio of MgCl₂ : Cl^- = 1:2

moles of chloride ions = 0.01263 * 2 =0.02526 moles of Chloride ions.

Molarity /concentration of Cl^- is

$M = 0.02562 mol / 2.0L = 0.0126 M$ Or 0.0126 mol/L