Question

What volume of a 8.50M Na2CO3 solution would be required to prepare 220.00mL of a 8.20%w/v Na2CO3 solution?

Answer 1

First, calculate what you need. 8.2% w/v means 8.2 g Na2CO3/100. You want 220 mL which means 8.2 g x (220 mL/100 mL) = about 18 g Na2CO3.

Now see what you have.
8.5M Na2CO3 means 8.5 moles/L soln and 8.5 moles is 8.5 x molar mass Na2CO3 = approximately 900 g (but you need to do it more accurately).

THEN we can calculate how much we need to take.
We want 1000 mL x (18/900) = ? mL.