Calculate the pressure in an evacuated 250. mL container at 0.00 degrees C if 3.20 cm^3 of liquid oxygen evaporates. Assume that the liquid oxygen has a density of 1.118 g/cm^3

I understand that the density and volume can be used to find the mass in grams. However, I'm confused about how the 3.20 cm^3 of oxygen that evaporates translates into the problem. Thanks in advance!

1.118 g/cc x 3.20 cc = ?grams O2.

?grams O2/molar mass O2 = n O2.
Then use n in PV = nRT and solve for P.

To calculate the pressure in the container, we need to use the ideal gas law equation:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant (usually given as 0.0821 L·atm/(mol·K))
T = Temperature in Kelvin

First, we need to calculate the number of moles of gas present in the container. To do this, we need to find the mass of the evaporated liquid oxygen.

The volume of the gas is given as 250 mL. However, we need to convert it to liters since the ideal gas constant is given in liters.

250 mL * (1 L / 1000 mL) = 0.25 L

The volume of the evaporated liquid oxygen is given as 3.20 cm^3. To use this information, we need to consider that the liquid oxygen has a density of 1.118 g/cm^3.

First, we need to calculate the mass of the liquid oxygen using the volume and density:

Mass = Volume * Density = 3.20 cm^3 * 1.118 g/cm^3 = 3.57 g

Now, we can calculate the number of moles of gas using the mass and the molar mass of oxygen (32 g/mol):

n = Mass / Molar mass = 3.57 g / 32 g/mol = 0.1115 mol

Next, we need to convert the temperature from degrees Celsius to Kelvin. To do this, we add 273.15 to the temperature:

Temperature in Kelvin = 0.00°C + 273.15 = 273.15 K

Now that we have all the necessary values, we can substitute them into the ideal gas law equation:

P * 0.25 L = 0.1115 mol * 0.0821 L·atm/(mol·K) * 273.15 K

P = (0.1115 * 0.0821 * 273.15) / 0.25
P = 1.075 atm

Therefore, the pressure in the evacuated container at 0.00 degrees Celsius is approximately 1.075 atm.