Consider a gas that has a molar mass of 100.21 g/mol. What mass of this gas would occupy a 150 mL container under a pressure of 1.00 atmosphere and a temperature of 25 degrees centegrade?
To find the mass of the gas, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Let's start by converting the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25°C + 273.15 = 298.15 K
The pressure is given as 1.00 atmosphere, which we can keep as is.
The volume is given as 150 mL, which we need to convert to liters:
V(L) = V(mL) / 1000
V(L) = 150 mL / 1000 = 0.150 L
Now, we can rearrange the ideal gas law equation and solve for the number of moles (n):
n = (PV) / (RT)
Substituting the given values into the equation:
n = (1.00 atm * 0.150 L) / (0.0821 L*atm/mol*K * 298.15 K)
n = 0.015 L*atm / (0.0821 L*atm/mol*K * 298.15 K)
Now, we can calculate the number of moles:
n = 0.015 L*atm / (24.466 L*atm/mol*K)
n = 0.000613 mol
Finally, we can calculate the mass of the gas using its molar mass:
Mass = molar mass * number of moles
Mass = 100.21 g/mol * 0.000613 mol
Calculating the mass:
Mass ≈ 0.0614 g
Therefore, the mass of the gas that would occupy a 150 mL container under a pressure of 1.00 atmosphere and a temperature of 25 degrees Celsius is approximately 0.0614 grams.