89.3k views
3 votes
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.)

User Lsaudon
by
7.9k points

1 Answer

3 votes

Answer:

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

Step-by-step 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.

User Joseph Gagnon
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories