An experiment was conducted using a 2M KCl solution

Question

An experiment was conducted using a 2M KCl stock solution. There was 1L of stock KCl solution and 480ml of distilled water at the beginning of the experiment, and there is only 125 ml of stock and 45ml of water remaining. What was the final volume of the dilution? What was the final concentration?

Answer 1

Initial volume 2M KCl = 1 L = 1000 ml

Final volume 2M KCl = 125 ml

Volume of 2M KCl used = 1000 ml - 125 ml = 875 ml

 

Initial volume H2O = 480 ml

Final volume H2O = 45 ml

Volume H2O used = 480 ml - 45 ml = 435 ml

 

Final volume of solution = 875 ml KCl + 435 ml H2O = 1,310 ml = 1.31 L

Final concentration of KCl:  (875 ml)(2 M) = (1310 ml)(x M) and x = 1.34 M KCl

Since 2 M has only 1 significant figure, you should report the final concentration of KCl as 1 M