To what volume shuould you dilute 125 mL of an 8.00 mL CuCl2 solution so that 50.0 mL of the diluted solution contains 5.9 g CuCl2?
M= mol solute / vol solution (L)
m = mol solute / mass solvent (kg)
mole percent = mol solute / mol solute + solvent
The Attempt at a Solution
8.00 M = mol CuCl2 / .125 L
1 mol CuCl2 x (65.55 + 2(35.45)g / 1 mol CuCl2 = 134.45 g CuCl2
5.9 g CuCl2 (final) x (1 mol CuCl2) / (65.55 +2(35.45)g = .4388 mol CuCl2
8.00 M = .44 mol CuCl2 --> .055 L solution???????????
I don't know how to use these formulas to find the diluted amount using the old amount? Do I subtract the grams CuCl2 to the new Molarity and find the tot vol? The answer is not the same as the back of the book.
THANKS SO MUCH!