Asked by Cindy

How much solute is required to make 600mL of a 0.4 M solution of Tris buffer? FW of Tris Buffer= 121.1g


Work: Molarity equation= moles of solute/ liters of solution

Step 1: 121.1 g / 121.14 g/mol TB = 0.9996 mol TB


Step 2: Molarity Eq'n= moles of solute/ liters of sol'n ... so, 0.9996/0.6 L = 1.666 mole (M) of solute

OR isn't it just ... 0.4 M sol'n = moles of solute/0.6 L = 0.24 mol of solute ?

Answered by bobpursley
Wow, did you go a journey.

it is the "just...", .24mol of solute, and you can convert that to mass
.24mol*121g=29g
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