The molarity of Cl^- in an aqueous solution which has w/v 2% NaCl, 4% CaCl₂ and 6% NH₄Cl will be

a) 0.342

b) 0.721

c) 1.12

d) 2.18

Let us suppose that the volume of the solution = 100 mL

Mass of NaCl in solution = 2 g

Mass of CaCl₂ in solution = 4 g

Mass of NH₄Cl in solution = 6 g

1 mole (58.44 g) of NaCl will give 1 mole (35.5 g) of Cl^- ion.

Therefore, 2 g of NaCl will give (35.5 x 2) / 58.44 = 1.21 g of Cl^- ion.

1 mole (111 g) of CaCl₂ will give 2 mole (35.5 x 2 g = 70 g) of Cl^- ion.

Therefore, 4 g of CaCl₂ will give (70 x 4) / 111 = 2.52 g of Cl^- ion.

1 mole (53.5 g) of NH₄Cl will give 1 mole (35.5 g) of Cl^- ion.

Therefore, 6 g of NH₄Cl will give (35.5 x 6) / 53.5 = 3.98 g of Cl^- ion.

Total mass of Cl^- ion in solution = 1.21+2.52+3.98 = 7.71 g

Molarity = (weight of solute x 1000) / (Molar mass of solute x volume of solution)

Molarity = (7.71 x 1000) / (35.5 x 100) = 2.17 M