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.