You have a 0.5M stock solution of NaCl (formula weight: 58.4 g/mole), a 0.15M stock solution of glucose (formula weight: 180.2g/mole), and a bottle of solid Tris base (formula weight: 121.1g/mole). How would you prepare (be specific) 25 mL of a single solution containing 10mM Tris, 15mM glucose, and 5mM NaCl.
You have a 0.5M stock solution of NaCl (formula weight: 58.4 g/mole), a 0.15M stock solution of glucose (formula weight: 180.2g/mole), and a bottle of solid Tris base (formula weight: 121.1g/mole). How would you prepare (be specific) 25 mL of a single solution containing 10mM Tris, 15mM glucose, and 5mM NaCl.
Introduction
The standard concentration of the solution is usually determined in "M", which is mole per liter of the solution.
Which is the unit of the molarity which is defined as the number of the moles of the solute dissolved in 1Litre of the solution.
The mathematical expression to calculate the molarity is as follows:
M=Moles of the solute / Volume of the solution in Litre
The dilution of a particular amount of solution will result in the final concentration given by
Final concentration=Initial concentration×Volume taken/Total volume after dilution.
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