Question

33.

What is the molarity of SO4 ion in aqueous

solution that contain 34.2 ppm of Al,(SO), ?

(Assume complete dissociation and density of

solution 1 g/mL)

(a) 3 x 104 M

(b) 2 x 10-4M

(c) 104M

(d) None of these

Answer 1

Answer:

1.Molarity=34.2×10^-4M

2.Thus molarity of Al2SO4is34.2*10^-4M

3. The molarity of SO4^2-=34.2×10^-4M

Answer 2

The molarity of sulfate ions in aqueous solution is equal to 3×10^-4 M.

• 1 ppm is equivalent to one milligram of solute present per kg of the solution.
• So, 34.2 mg of Al2(SO4)3 is present per kg of the solution.
• Since density is equal to 1 g/ml , 1 kg of solution occupies 1 L volume.
• Molar mass of solute is equal to 342 g/mole.
• moles of solute present is equal to 10^-4.
• Moles of sulfate ions is equal to 3 times the moles of solute (i.e, 3×10^-4 moles).
• As 3×10^-4 moles are present per L of the solution , molarity = 3×10^-4 M