posted by hanan .
A sample of air contains 32.7 µg/m3
of beryllium dust. How many atoms of beryllium are
present in a room with dimensions of 8 feet by
13.6 feet by 15.6 feet?
Answer in units of atoms
Compute the room volume in ft^3 and convert it to cubic meters.
8*13.6*15.6 = 1797 ft^3 = 50.9 m^3
With 32.7*10^-6 g per m^3, there are 1.66*10^-3 gram of Be dust in the room.
Convert that to moles using the atomic weight of Be (9.0 g/mol), and then use Avogadro's number for the number of atoms.