How would you make 2.5 liters of an aqueous solution containing 150mM Tris (the MW of Tris is 120g/mole) ? (Hint: Do the necessary

calculations AND then describe how the solution would be made.)

45 g of the solid Tris will be dissolved in 2.5 liters of water.

Explanation:

Recall that:

Number of moles = molarity x volume

Hence, number of moles of Tris present in 2.5 liters, 150 mM solution:

= 150/1000 x 2.5 = 0.375 moles

Also, recall that:

No of moles of substance = mass/molar mass.

Hence, mass of 0.375 moles substance:

= no of moles of the substance x molar mass of the substance.

= 0.375 x 120 = 45 g.

Therefore, in order to prepare 2.5 liters, 150 mM of an aqueous solution of Tris, 45 g of the solid Tris will be dissolved in 2.5 liters of water.