Using the formula weight of tris (fw = 121.1), how much (to the nearest 0.01 g) is needed to prepare 50 ml of a 0.50 m solution?

3.03 g

Solution:

Data Given:

Molarity = 0.50 mol. L⁻¹

M. Mass = 121.1 g. mol⁻¹

Volume = 50 mL = 0.05 L

Step 1: Calculate moles as,

Molarity = Moles : Volume

Solving for Moles,

Moles = Molarity * Volume

Putting values,

Moles = 0.50 mol. L⁻¹ * 0.05 L

Moles = 0.025 mol

Step 2: Calculate Mass of Tris as,

Moles = Mass : M. Mass

Solving for Mass,

Mass = Moles * M. Mass

Putting values,

Mass = 0.025 mol * 121.1 g. mol⁻¹

Mass = 3.03 g