To calculate the number of Cl2 (chlorine) molecules present in a given volume, we need to use the ideal gas law and convert the volume to the appropriate units.
The ideal gas law states:
PV = nRT
Where: P = pressure of the gas V = volume of the gas n = number of moles of gas R = ideal gas constant T = temperature of the gas in Kelvin
Assuming standard temperature and pressure (STP), the pressure (P) is 1 atmosphere, and the temperature (T) is 273.15 Kelvin.
The ideal gas constant (R) is 0.0821 L·atm/(mol·K).
To convert the given volume of 200 cm^3 to liters (L), we divide by 1000 since 1 L = 1000 cm^3.
V = 200 cm^3 / 1000 = 0.2 L
Now, we can rearrange the ideal gas law to solve for the number of moles (n):
n = PV / RT
Plugging in the values:
n = (1 atm) * (0.2 L) / (0.0821 L·atm/(mol·K)) * (273.15 K)
Simplifying the expression:
n = 0.0082 mol
Since one molecule of Cl2 consists of two chlorine atoms, the number of Cl2 molecules is the same as the number of chlorine atoms.
Therefore, in 0.0082 mol of Cl2, there are 0.0082 mol * (6.022 x 10^23 molecules/mol) = 4.9444 x 10^21 Cl2 molecules (or chlorine atoms).
So, there are approximately 4.9444 x 10^21 Cl2 molecules (or chlorine atoms) in a 200 cm^3 volume.