Question

5.4263 g of Na2CrO4 (MW = 161.97 g/mol) is dissolved in 100.0 mL of water. Assuming the solution has a density of 1.00 g/mL, what is the concentration of Na (MW = 22.9898 g/mol) in the solution in units of

a. molarity

b. parts per thousand (ppt)

c. 50.0 mL of the solution is then diluted to a final volume of 1000.0 mL. What is the concentration of Na in the diluted solution in units of parts per million (ppm)?

Answer 1

Moles of Na₂CrO₄ = 5.4263/161.97 = 0.0335 moles

Now, as and ​Na₂CrO₄ Na are in 1:2 molar relation, the moles of Na will be twice the moles of Na₂CrO₄.

Moles of Na = 0.0670 moles

(a) Molarity = Moles/Volume = 0.0670/0.1

Molarity = 0.67 mol/L

(b) ppt = 1ng/L

Concentration = (0.67 x Molar mass of Na) g/L

Concentration = 15.403166 g/L = 15.403 X 10⁹ ng/L = 15.403 X 10⁹ ppt

(c) Now, to find the molarity of the diluted solution, we use M₁V₁ = M₂V₂

0.67 X 0.05 = 1 X M₂, M₂ = 0.0335 mol/L (Note : 1 ppm = 1mg/L

As earlier, Concentration = 0.7701583 g/L = 770.158 mg/L = 770.158 ppm.