Halogenoalkane question

hello, i am stuck with a question and i will post it below shortly as clicking ask a question button does not let me upload the picture as i wanted

Reply 1

Batman2k1

small amount of halothane CF CH BR CL with mass 3.948x10^-4 kg escapes from room of volume 150m^3. Calculate the number of halothane molcules in a volume of 500cm^3. Avodgadro constant is 6.02x10^23

Answer: Mr of CF3CHBrCl  197.4
Moles of halothane  3.948 ´10 –4 ´10 3 197.4  0.002
Number of molecules  0.002 x 6.02 ^ 1023 = 1.204x10^21
Molecules in 500 cm3  (1.204 ´10^21 ´ 500^10 –6 ) 150 = 4.01 x10^15

I understand every bit expect the molecules in 500cm3 can someone explain this 2 me thanks.