Question

Two liters of a standard Zn^2+ solution is to be made by dissolving 0.5265g solid zinc nitrate hexahydrade in 100mL of 0.100 MHNO_3. The solution was made up to two liters in volume. What is the concentration of the Zn (ppm) standard solution?

Answer 1

Mass of Zinc nitrate hexahydrate, Zn(NO₃)₂.6H₂O = 0.5265 gram

Molecular mass of Zn(NO₃)₂.6H₂O = 297.39 gram / mol

Number of moles = mass / molecular mass

                        = 0.5265 gram / (297.39 gram / mol) = 0.00177 moles

So, number of moles of Zn²⁺ in given sample = number of moles of Zn(NO₃)₂.6H₂O because each mole of this compound has 1 mole of Zn.

Thus, number of moles of Zn²⁺ = 0.00177 moles

Mass of Zn²⁺ = moles of Zn x atomic mass

                        = 0.00177 moles x 65.38 gram per mol = 0.1157489 gram

                        = 115.75 mg                           [1 gram =1000 mg]

[Zn²⁺] in ppm = mass of Zn in mg / volume of solution in L

                        = 115.75 mg / 2 L                                           [1 ppm = 1 mg/ L]

                        = 57.87 ppm = 58 ppm (approx.)