100. g CCl4* (1 mol CCl4/ 153.8 g CCl4)* (6.02*10^23 CCl4 molecules/ 1 mol CCl4)= 3.91*10^23 CCl4 molecules.
(Note that the units cancel out so you get the answer)
Hope this helps~
Number of moles of the gas, Temperature and the volume of the gas.
<span>Divide the number of grams present in the sample by copper's gram atomic weight to find the number of gram atomic weights present. Then multiply that result by Avogadro's Number: 6.022137 x 10^23 atoms/gram atomic weight.1,200 g/(63.54 g/gram atomic weight) ? 18.885741 gram-atomic weights. Hope this helps. </span>
Hi , the answer is 5.678 x 10^3.
