posted by Becca .
When 26 mL of a 120 mL stock solution containing 3.23 M CaF2 is diluted to 35 mL, how many moles of F- are in the new solution
M CaF2 in new soln = 3.23M x (26/35) = ?M
which isn't much change due to dilution.
moles CaF2 = ?M x L = ?M x 0.026L = ?
You have two F ions/mol CaF2; therefore multiply mol CaF2 x 2 = moles F ions.
By the way, frankly I don' think you can make 3.23 M soln of CaF2. It just isn't that soluble. In fact, you can't even come close to 3.23 M CaF2 unless you add something else to the water.