Question

Use the following data provided to calculate the mass thats needed for the solution.

DDCNa.3H2O (FW=225.31) 50 ml of 2% w/v DDCNa solution in methanol

NaAc.3H2O (FW=136.08) 50 ml of 5% w/v NaAc solution in DI water

CuSO4 .5H2O (FW=249.68) 100 ml of 300 ppm Cu2+ solution in 1% HNO3

Answer 1

Ans. #1. Given, DDCNa content = 2 % (w/v)

2% w/v means that there is 2.0 g of DDCNa in 100.0 mL of the solution.

So,

            Mass of DDCNa = 2 % (w / v) of 50.0 mL

                                                = (2.0 g / 100.0 mL) x 50.0 mL

                                                = 1.0 g

#2. 5% w/v means that there is 5.0 g of NaAc in 100.0 mL of the solution.

So,

            Mass of NaAc = 5 % (w / v) of 50.0 mL

                                                = (5.0 g / 100.0 mL) x 50.0 mL

                                                = 2.5 g

#3. 1 ppm = 1 mg of solute per 1.0 L of solution

So,

            300 ppm Cu²⁺ = 300.0 mg Cu²⁺ per liter of solution

                                    = 300.0 mg Cu²⁺ / L

Given,

            Volume of solution = 100.0 mL = 0.100 L                        ; [1 L = 1000 mL]

Now,

            Mass of Cu²⁺ = [Cu²⁺] x Desired volume of solution in liters

                                    = (300.0 mg / L) x 0.100 L

                                    = 30.0 mg

                                    = 0.030 g