If a 14.05g sample of a gas occupies 10.0L at STP, what is the molar mass of the gas at 125 degrees Celsius?
To determine the molar mass of the gas at 125 degrees Celsius, we can use the ideal gas law. The ideal gas law equation is given as:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in liters)
n = Moles of gas
R = Gas constant (0.0821 atm·L/mol·K)
T = Temperature (in Kelvin)
At standard temperature and pressure (STP), conditions are defined as 1 atmosphere of pressure and 273.15 Kelvin temperature.
Given:
Mass of gas (m) = 14.05 grams
Volume (V) = 10.0 Liters
Temperature (T) = 125 degrees Celsius (convert to Kelvin)
To convert Celsius to Kelvin, we use the following equation:
T(K) = T(°C) + 273.15
Let's calculate:
T = 125°C + 273.15 = 398.15 Kelvin
Now, we can rearrange the ideal gas law equation to solve for moles (n):
n = PV / RT
Plug in the values:
n = (1 atm) * (10.0 L) / (0.0821 atm·L/mol·K) * (398.15 K)
Simplifying:
n = 0.122 mol
Now we have the number of moles (n), and we know the mass (m) = 14.05 grams. To find the molar mass (M) of the gas, we use the formula:
Molar mass (M) = mass (m) / moles (n)
M = 14.05 g / 0.122 mol
Simplifying:
M ≈ 115 g/mol
Therefore, the molar mass of the gas at 125 degrees Celsius is approximately 115 g/mol.