To find the volume occupied by 10.0 g of Cl₂ gas at 25 °C and 0.900 atm, we can use the ideal gas law, which is represented by the equation:
PV = nRT
Where:
P = Pressure of the gas (in atm)
V = Volume of the gas (in liters)
n = Number of moles of gas
R = Ideal gas constant (0.0821 L.atm/(mol.K))
T = Temperature of the gas (in Kelvin)
First, we need to calculate the number of moles (n) of Cl₂ gas. To do that, we can use the molar mass of Cl₂, which is the sum of the molar masses of two chlorine atoms:
Molar mass of Cl₂ = 2 * Molar mass of Cl = 2 * 35.45 g/mol ≈ 70.90 g/mol
Now, we can find the number of moles (n):
n = Mass of Cl₂ / Molar mass of Cl₂
n = 10.0 g / 70.90 g/mol ≈ 0.141 mol
Next, we need to convert the temperature from Celsius to Kelvin:
Temperature in Kelvin (T) = 25 °C + 273 ≈ 298 K
Now, we have all the necessary values to calculate the volume (V) using the ideal gas law:
V = nRT / P
V = (0.141 mol) * (0.0821 L.atm/(mol.K)) * (298 K) / 0.900 atm
V ≈ 3.83 L
Answer:
Therefore, the volume occupied by 10.0 g of Cl₂ gas at 25 °C and 0.900 atm is approximately 3.83 liters.