Given that on average, an atom of gold weighs 3.27×10^−22grams, how much
does an Avagadro’s number of gold atoms weigh (take Avagadro’s number to be 6.0221415×10^23).
just multiply the mass of each atom by the number of atoms:
3.27*10^-22 * 6.0221415*10^23 = 19.6924*10^1 = 196.924 g
check: mol wt of Au is 196.97. Pretty close
To find out how much an Avogadro's number of gold atoms weighs, we need to multiply the weight of a single gold atom by Avogadro's number.
Given that an atom of gold weighs 3.27×10^−22 grams and Avogadro's number is 6.0221415×10^23, we can calculate:
Weight of Avogadro's number of gold atoms = (Weight of a single gold atom) * (Avogadro's number)
Weight of Avogadro's number of gold atoms = (3.27×10^−22 grams) * (6.0221415×10^23 atoms)
To multiply the scientific notation, you need to multiply the coefficients and add the exponents:
Weight of Avogadro's number of gold atoms = 3.27 * 6.0221415 * 10^(-22 + 23) grams
Simplifying further:
Weight of Avogadro's number of gold atoms = 19.709752545 * 10^1 grams
Finally, we can convert the exponential form to decimal form:
Weight of Avogadro's number of gold atoms = 19.709752545 * 10 grams
Weight of Avogadro's number of gold atoms = 197.09752545 grams
Therefore, an Avogadro's number of gold atoms weighs approximately 197.09752545 grams.