Using the formula weight of tris (fw = 121.1), how much (to the nearest 0.01 g) is needed to prepare 50 ml of a 0.50 m solution?

              3.03 g

Solution:

Data Given:

                  Molarity  =  0.50 mol.L⁻¹

                  M.Mass  =  121.1 g.mol⁻¹

                  Volume  =  50 mL  =  0.05 L

Step 1: Calculate moles as,

                            Molarity  =  Moles ÷ Volume

Solving for Moles,

                            Moles  =  Molarity × Volume

Putting values,

                            Moles  =  0.50 mol.L⁻¹ × 0.05 L

                            Moles =  0.025 mol

Step 2: Calculate Mass of Tris as,

                            Moles  =  Mass ÷ M.Mass

Solving for Mass,

                            Mass  =  Moles × M.Mass

Putting values,

                            Mass  =  0.025 mol × 121.1 g.mol⁻¹

                            Mass =  3.03 g

pie chart because it is in percents ( % )

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.