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## Homework Statement

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.

## Homework Equations

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.