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Tris has a molecular weight of 121 g/mol. How many grams of Tris would you need to make 100 mL of a 100 mM solution of Tris

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Answer:

1.21 g of Tris

Step-by-step explanation:

Our solution if made of a solute named Tris

Molecular weight of Tris is 121 g/mol

[Tris] = 100 mM

This is the concentration of solution:

(100 mmoles of Tris in 1 mL of solution) . 1000

Notice that mM = M . 1000 We convert from mM to M

100 mM . 1 M / 1000 mM = 0.1 M

M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris

0.1 M = mmoles of Tris / 100 mL

mmoles of Tris = 100 mL . 0.1 M → 10 mmoles

We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol

And now we determine the mass of solute, by molecular weight

0.010 mol . 121 g /mol = 1.21 g

User Ray K
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