Question

Two solutions will be prepared with the mixture of distilled water and stock solution of KMnO4 (10 mM). Give the correct volumes of stock solution and water for both of the solutions listed below. The final volume in each tube should be 2 ml

Tube A: 3 mM KMnO4
Tube B: 8 mM KMnO4
Answer format :(Show correct calculation 2 pts, correct answer 2 points) Tube A should be made with [X] mL of 10 mM KMnO4 stock and [Y] mL of distilled water Tube B should be made with [X] mL of 10 mM KMnO4 stock and [Y] mL of distilled water

Answer 1

See the answer below

Step-by-step explanation:

Recall that the law of dilution states that the number of moles before dilution must be equal to the number of moles after dilution.

Mathematically,

molarity x number of moles before dilution = molarity x number of moles after dilution.

For solution A: final molarity = 3 mM, final volume = 2mL, initial molarity of KMnO4 = 10 mM

Applying the equation:

10 x initial volume = 3 x 2

initial volume = 6/10 = 0.6

Hence, Tube A should be made with 0.6 mL of 10 mM KMnO4 stock and 1.4 mL of distilled water to give a solution of 2 mL 3 mM KMnO4.

For solution B:final molarity = 8 mM, final volume = 2 mL, initial molarity = 10 mM

10 x initial volume = 8 x 2

initial volume = 16/10 = 1.6

Hence, Tube B should be made with 1.6 mL of 10 mM KMnO4 stock and 0.4 mL of distilled water to give a solution of 2 mL 8mM KMnO4.