A golf O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed from a 100-mm length of 1.5-mm-diameter wire. Calculate the number of gold atoms in the O-ring.
Volume of cylinder = PI*h*r^2
The Attempt at a Solution
I got the volume of the cylinder:
PI*100*.75^2 = 176.7
I used the formula m = D*V:
density gold = 19.28 g/cm^3
amu gold = 196.97 grams
19.28*176.7 = 3406.776
I then multiplied by avogadro's number over amu.
3406.776 *(6.02x10^23 / 196.97) = 1.04x10^25 atoms
The answer should be 10.4x10^21 atoms. I'm thinking I messed up some unit conversion. I am unsure of where.