How many grams of gas are present in a sample that has a molar mass of 44g/mol and occupies a 1.8-L container at 108 kPa and 26.7C?

Answer 1

To calculate the mass of the gas present in the container, we can use the ideal gas law equation:

PV = nRT

Where:
P = 108 kPa = 108000 Pa
V = 1.8 L = 0.0018 m^3 (since 1 L = 0.001 m^3)
R = 8.31 J/(mol·K) (universal gas constant)
T = 26.7°C = 26.7 + 273.15 = 299.85 K (temperature in Kelvin)

First, we need to calculate the number of moles of gas present in the container:

n = PV/(RT)
n = (108000 Pa)(0.0018 m^3)/(8.31 J/(mol·K) × 299.85 K)
n ≈ 0.0887 moles

Now, we can calculate the mass of the gas using the molar mass:

Mass = n × molar mass
Mass = 0.0887 moles x 44 g/mol
Mass ≈ 3.90 grams

Therefore, there are approximately 3.90 grams of gas present in the 1.8-L container at 108 kPa and 26.7°C.