Question

a solution was prepared by dissolving 5.76g of KCL*MgCl2*6H2O (277.85g/mol) in sufficient water to give 2.000L. Calculate

A)molar analytical concentration of KCL*MgCl2 in this solution
B)molar concentration of Mg2+
C) molar concentration of Cl-

Answer 1

Given mass of KCl*MgCl₂*6H₂O , m = 5.76 g
Molar mass ofKCl*MgCl₂*6H₂O , W = 277.85 g/molVolume of the solution , V = 2.00 LWe know that molarity , M = ( mass/Molar mass) / volume of solution in L= ( m / W ) / V= ( 5.76 / 277.85 ) / 2.00 M= 0.01036 MKCl*MgCl₂*6H₂O ----> Mg ²⁺ + 2Cl^- +K⁺ + Cl^- + 6H₂O----------------------from MgCl₂ from KClKCl*MgCl₂*6H₂O ----> Mg ²⁺ + 3Cl^- +K⁺ + 6H₂O1 mole of KCl*MgCl₂*6H₂O produces 1 mole of Mg ²⁺ &3 moles of Cl^-So molarity of Mg ²⁺ = 1 x molarity of KCl*MgCl₂*6H₂O= 1 x 0.01036 M= 0.01036 M1 mole of KCl*MgCl₂*6H₂O produces3 moles of Cl^-So molarity of Cl^- =3 x molarity of KCl*MgCl₂*6H₂O=3 x 0.01036 M= 0.0311 M