Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Huh
Lv 6
Huh asked in Science & MathematicsChemistry · 2 years ago

What is the derivation for Avagadro’s number?

4 Answers

Relevance
  • 2 years ago

    You want to know how was this number derived ( That is calculated). You do not want a definition of this number.

    Here is an explanation I found from Scientific American:

    Accurate determinations of Avogadro’s number require the measurement of a single quantity on both the atomic and macroscopic scales using the same unit of measurement. This became possible for the first time when American physicist Robert Millikan measured the charge on an electron. The charge on a mole of electrons had been known for some time and is the constant called the Faraday. The best estimate of the value of a Faraday, according to the National Institute of Standards and Technology (NIST), is 96,485.3383 coulombs per mole of electrons. The best estimate of the charge on an electron based on modern experiments is 1.60217653 x 10-19 coulombs per electron. If you divide the charge on a mole of electrons by the charge on a single electron you obtain a value of Avogadro’s number of 6.02214154 x 10²³ particles per mole.

  • 2 years ago

    It's 12 grams divided by the actual mass (in grams) of a single atom of the carbon-12 isotope.

  • Anonymous
    2 years ago

    it's called avogadro's number. See morningfox's answer

  • 2 years ago

    How far back do you want to go? In 1909, it was the number of molecules in 32 grams of oxygen. Then in 1971, it was defined as the number of atoms in 12 grams of carbon-12. Then in 2017 (effective on May 2019) it was redefined as 6.02214076 x 10^23 (exactly).

Still have questions? Get your answers by asking now.