Question

You have a 5M stock solution of NaCl (Formula Weight: 58.44g/mole), a 0.25M stock solution of glucose (Formula Weight; 180.156g/mole), and a bottle of solid Tris base (Formula Weight: 121.1g/mole). How would prepare (be specific) 250mL of a single solution containing 150mM Tris, 25mM glucose, and 150mM NaCl. g

Answer 1

4.54g of Tris base,25mL of the 0.25M stock solution of glucose and 7.5mL of the 5M stock solution of NaCl must be added and complete the volume in a volumetric flask to 250.0mL

Step-by-step explanation:

To prepare the single solution we need to find the moles of each solute (Tris, glucose and NaCl) from the stock solutions anf the solid:

Moles Tris:

0.250L *(0.150mol / L) = 0.0375moles Tris * (121.1g/mol) = 4.54g of Tris base must be added

Moles glucose:

0.250L * (0.025mol/L) = 6.25x10⁻³mol glucose * (1L / 0.25mol) = 0.025L = 25mL of the 0.25M stock solution of glucose must be added

Moles NaCl:

0.250L * (0.150mol / L) = 0.0375mol NaCl * (1L / 5mol) = 0.0075L =

7.5mL of the 5M stock solution of NaCl

You must add:

4.54g of Tris base,25mL of the 0.25M stock solution of glucose and 7.5mL of the 5M stock solution of NaCl must be added and complete the volume in a volumetric flask to 250.0mL