1.When 20 mL of a 145 mL stock solution containing 2.2 M BaF2 is diluted to 70 mL, what is the concentration of F- in the new solution?
anw:1.25
2.
Solutions and Dilutions
2008-03-30 5:14 pm
回答 (1)
2008-03-30 8:13 pm
✔ 最佳答案
Before dilution:M1 = 2.2 M
V1 = 20 mL
After dilution
M2 = ?
V2 = 70 mL
M1V1 = M2V2
2.2 x 20 = M2 x 70
M2 = 2.2 x 20 / 70 = 0.628 M
Concentration of MF2 after dilution = 0.628 M
1 mol of MF2 gives 2 mol of F-.
= Concentration of F- after dilution
= 2 x 0.628
= 1.26 M
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