The preparation of homeopathic \"remedies\" involves the repeated dilution of so

Question

The preparation of homeopathic "remedies" involves the repeated dilution of solutions containing an active ingredient such as arsenic trioxide (As2O3). Suppose one begins with 57.1 g of arsenic trioxide dissolved in water, and repeatedly dilutes the solution with pure water, each dilution reducing the amount of arsenic trioxide remaining in the solution by a factor of 100. Assuming perfect mixing at each dilution, what is the maximum number of dilutions one may perform so that at least one molecule of arsenic trioxide remains in the diluted solution? For comparison, homeopathic "remedies" are commonly diluted 15 or even 30 times.

Explanation / Answer

first find the molecular mass of As2 O3 to determine how many moles are in 6.7 g and therefore how many molecules--this has a molecular mass of 197.8 g/mole

therefore there are 6.7 g/197.8g/mol = 0.0338 moles and therefore 0.39xAvogadro's numbers of molecules

=0.03338mol x 6.023x10^23molecules/mol = 2.04 x10^22 molecules

each dilution reduces the number of molecules by a factor of 10^2

n dilutions reduces the number of molecules by 10^2n meaning if we dilute 5 times, we have 1/10^10 times as many particles as we started with

therefore, to have at least one molecule remaining, we cannot dilute by more than a factor of 10^22, or we can not have more than 11 dilutions

the problem shows that homeopathic remedies are so diluted, that there is not even one molecule of the supposedly healing agent left in the solution

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