Question

How many atoms of ca are present in 8.6 mg? the atomic weight of ca is 40.08 amu?

Answer 1

One mole of any element is Avogadro's number, which is 6.02 * 10^23 atoms. One mole of Calcium is equal to its atomic weight in grams, 1 mole Ca = 40.08 x 1000 mg = 6.02 * 10^23 atoms. We have 8.6mg of Ca which means 8.6mg / 40,080mg = .00021457 of a mole of calcium. Now multiply that by Avogadro's number which gives number of atoms = 0.00021457 * 6.02 * 10^23 atoms. So the answer is 1.292 * 10^20 atoms of Ca.