Question

i need to make a 2.0 mM NaPhytate ( 923.8 g/mol) solution in 100mM NaAcetate i need to make a 8.4 ml solution

Answer 1

Ans. It’s assumed that-

            I. Na-phytate and Na-acetate do not react/interfere with each other.

            II. Addition of Na-phytate to Na-acetate doesn’t change the volume.

            III. Both the solutes are completely soluble in water.

# Given, volume of solution = 8.4 mL = 0.0084 L

# Given, [Na-phytate] = 2.0 mM = 0.002 M                                 ; [1 mM = 0.001 M]

Now,

            Required moles of Na-phytate = [Na-phytate] x vol. of solution in liters

                                                            = 0.002 M x 0.0084 L

                                                            = 0.0000168 mol

            Required mass of Na-phytate = Required moles x Molar mass

                                                            = 0.0000168 mol x (923.8 g/ mol)

                                                            = 0.01551984 g = 0.01552 g

                                                            = 15.5198 mg

# Given, [Na-acetate] = 100 mM = 0.100 M                                ; [1 mM = 0.001 M]

Now,

            Required moles of Na-acetate = 0.100 M x 0.0084 L = 0.00084 mol

            Required mass of Na-acetate = 0.00084 mol x (82.034388 g/ mol) = 0.0689 g

# Preparation of solution:

Option 1: If 100 mM Na-acetate is available: Take 0.01552 g Na-phytate in a clean 10.0 mL test tube. Add 8.4 mL of 100.0 mM Na-acetate solution (see, assumption 2). It is you desired solution.

# Option 2: If 100 mM Na-acetate is not available: Prepare 100.0 mM Na-acetate solution by dissolving 0.0689 g Na-acetate in 8.4 mL deionized water.

To the above solution, add 0.01552 g Na-phytate. Mix well. It is the desired solution.