A sample of Cl2 gas having a volume of 0.75 L at 20degrees C and 725mmHg has a mass equal to
To determine the mass of the Cl2 gas, we need to use the ideal gas law, which states:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in L)
n = number of moles of gas
R = ideal gas constant (0.0821 L•atm/(mol•K))
T = temperature of the gas (in Kelvin)
Let's convert the given values to the required units:
Temperature:
20 degrees Celsius + 273.15 = 293.15 K
Pressure:
725 mmHg = 725/760 atm (since 760 mmHg = 1 atm)
Convert to atm: 725/760 ≈ 0.954
Now, we can rearrange the ideal gas law equation to solve for the number of moles (n) of Cl2 gas:
n = PV / RT
Substituting the values:
n = (0.954 atm) x (0.75 L) / [(0.0821 L•atm/(mol•K)) x (293.15 K)]
n ≈ 0.0239 moles
Finally, to determine the mass of Cl2 gas, we'll use the molar mass:
Molar mass of Cl2 = 2 x atomic mass of Cl ≈ 2 x 35.45 g/mol ≈ 70.90 g/mol
Mass = n x molar mass
Mass ≈ 0.0239 moles x 70.90 g/mol ≈ 1.69 grams
Therefore, the mass of Cl2 gas is approximately 1.69 grams.
To determine the mass of Cl2 gas, we can use the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's convert the given values:
Volume (V) = 0.75 L
Temperature (T) = 20 degrees Celsius = 20 + 273.15 = 293.15 K
Pressure (P) = 725 mmHg = 725/760 atm (since 1 atm = 760 mmHg)
Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the values we have:
n = (725/760) atm * 0.75 L / (0.0821 L·atm/mol·K * 293.15 K)
Simplifying the equation:
n = (0.953) L·atm / (24.083 L·atm/mol·K)
n ≈ 0.0396 mol
Now that we have the number of moles of Cl2 gas, we can calculate its mass. The molar mass of Cl2 is 35.45 g/mol (34.47 g/mol for Cl + 0.98 g/mol for Cl):
Mass = number of moles * molar mass
Mass = (0.0396 mol) * (35.45 g/mol)
Mass ≈ 1.40 g
Therefore, the mass of the Cl2 gas sample is approximately 1.40 grams.