The following questions all refer to an aqueous 2.87 M calcium iodate stock solu
ID: 1002771 • Letter: T
Question
The following questions all refer to an aqueous 2.87 M calcium iodate stock solution. If 20.1 mL of the stock solution are diluted to 121 mL, what is the molarity (in mol/L) of the new solution? If 16.5 mL of the stock solution are diluted to obtain a 0.888 M calcium iodate solution, what is the volume (in mL) of the new solution? Determine the volume (in mL) of the original stock solution that must be used to prepare 470. mL of a 0.285 M calcium iodate solution. Determine the volume of water (in mL) that was added to 16.4 mL of the stock solution to prepare a 0.855 M calcium iodate solution.Explanation / Answer
Solution:- (1) To solve the dilution problems we use the dilution equation, M1V1 = M2V2
where M1 and V1 are initial molarity and volume. M2 and V2 are molarity and volume of diluted solution.
2.87 M x 20.1 ml = M2 x 121 ml
M2 = 2.87 M x 20.1 ml/121 ml = 0.477 M
So, the molarity of new solution is 0.477 M
(2) 2.87 M x 16.5 ml = 0.888 M x V2
V2 = 2.87 M x 16.5 ml /0.888 M = 53.3 ml
So, the volume of new solution is 53.3 ml
(3) 2.87 M x V1 = 0.285 M x 470. ml
V1 = 0.285 M x 470. ml / 2.87 M = 46.7 ml
So, the volume of the original stock solution is 46.7 ml.
(4) 2.87 M x 16.4 ml = 0.855 M x V2
V2 = 2.87 M x 16.4 ml / 0.855 M = 55.1 ml
so, volume of water added = 55.1 ml - 16.4 ml = 38.7 ml
So, 38.7 ml of water need to be added.
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