Practitioners of the branch of alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is the claim plausible? To explore this question, suppose that you prepare a solution of a supposedly active substance, X, with a molarity of 0.10 molL1 . Then you dilute 10. mL of that solution by doubling the volume, doubling it again, and so on, for 90 doublings in all. How many molecules of X will be present in 10. mL of the fi nal solution? Comment on the possible health benefi ts of the solution.

a. 4.865 × 10⁻⁷ molecules. b. The possible health benefits of the drug might be minimal or non-existent.

Explanation:

a. Since we have 0.10 mol/L and 10 mL of solution, the number of moles in 10 mL of solution is n = CV where C = concentration and V = Volume. C = 0.10 mol/L and V = 10 mL = 0.01 L.

So n = CV = 0.10 mol/L × 0.01 L = 0.001 mol.

Since the initial volume of the solution is 10 mL and doubled, it is 2 × 10 mL = 20 mL. It is doubled again to 2 × 20 mL = 40 mL. It is doubled again to 2 × 40 mL = 80 mL. Our third doubling is thus 80 mL = 2 × 2 × 2 × 10 mL = 2³ × 10 mL. So the volume for our nth doubling is V = 2ⁿ × 10 mL. So four our 90 th doubling, the volume is V = 2⁹⁰ × 10 mL = 2⁹⁰ × 10 mL

Since we have 0.001 mol present, the concentration in this new volume is thus C' = 0.001 mol/2⁹⁰ × 10 mL.

So the number of moles in 10 mL of this solution is thus n = 0.001 mol/(2⁹⁰ × 10 mL) × 10 mL = 0.001/2⁹⁰ mol

To find the number of molecules of X present we use, number of moles, n = number of molecules, N/Avogadro's number of molecules, N'

n = N/N'

N = nN'

= 0.001/2⁹⁰ mol × 6.022 × 10²³ molecules/mol

= 8.078 × 10⁻³¹ × 6.022 × 10²³

= 48.65 × 10⁻⁸

= 4.865 × 10⁻⁷ molecules.

b. Since the number of molecules of X present in 10 mL solution of the final diluted solution is 4.865 × 10⁻⁷ molecules, this shows that the number of molecules of X is very little so, the possible health benefits of the drug might be minimal or non-existent.