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S_A_V [24]
3 years ago
6

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

y
calculations AND then describe how the solution would be made.)
Biology
1 answer:
lisabon 2012 [21]3 years ago
6 0

Answer:

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

Explanation:

Recall that:

<em>Number of moles  = molarity x volume</em>

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:

<em>No of moles of substance = mass/molar mass.</em>

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.

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