3. Determine the density of chlorine gas at 22C and 1.00 atm pressure.
Use
PV=nRT= mass/molmass * RT
mass/Volume= Pressure*molmass/RT
Sorry, i don't know how to plug the numbers into this one.
To determine the density of chlorine gas at 22°C and 1.00 atm pressure, we need to use the ideal gas law equation and the molar mass of chlorine gas.
The ideal gas law equation is:
PV = nRT
Where:
P = pressure in atmospheres (1.00 atm in this case)
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 22 + 273.15
T(K) = 295.15 K
Since we are looking to find the density (mass/volume), we need to rearrange the ideal gas law equation to solve for density:
PV = nRT
PV = (m/M)RT (n = m/M, where m is mass and M is molar mass)
rho = (m/V) = P(M/RT)
Where:
rho = density
m = mass
V = volume
P = pressure
M = molar mass
R = ideal gas constant
T = temperature in Kelvin
Now, let's calculate the molar mass of chlorine gas (Cl2):
The molar mass of Cl (chlorine) = 35.45 g/mol
Since chlorine gas (Cl2) contains two chlorine atoms, the molar mass of Cl2 = 2 * 35.45 g/mol = 70.90 g/mol
Now we have all the values needed to calculate the density of chlorine gas:
P = 1.00 atm
M = 70.90 g/mol
R = 0.0821 L·atm/(mol·K)
T = 295.15 K
Plugging these values into the equation:
rho = (m/V) = P(M/RT)
rho = (1.00 atm) * (70.90 g/mol) / (0.0821 L·atm/(mol·K) * 295.15 K)
Simplifying the equation, we can see that the units cancel out:
rho = (1.00 atm) * (70.90 g/mol) / (0.0821 L·atm/(mol·K) * 295.15 K)
rho = (1.00 * 70.90) / (0.0821 * 295.15) g/L
rho ≈ 0.944 g/L
Therefore, the density of chlorine gas at 22°C and 1.00 atm pressure is approximately 0.944 g/L.