A 1.00 L vessel contains 5 g N2 and 1.5 g H2O at 25 degrees C. Calculate the mass of LIQUID water that would be present in the vessel.

The vapor pressure of H2O at 25C is 23.8 mm Hg. Plug that into PV = nRT and solve for n = mols H2O vapor. Subtract from mols in the 1.5g initially present to find mols H2O liquid.

To calculate the mass of liquid water present in the vessel, we first need to determine the amount of water vapor in the vessel and then subtract it from the initial mass of water.

To find the amount of water vapor, we can make use of the ideal gas law equation, which states:

PV = nRT

Where:
P is the pressure of the system
V is the volume of the system
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L · atm/(mol · K))
T is the temperature in Kelvin

Since we know the volume (1.00 L), temperature (25 degrees C or 298 K), and we assume the pressure is constant at atmospheric pressure, we can solve for the number of moles of water vapor.

Now that we have the number of moles of water vapor, we can calculate its mass using the molar mass of water, which is 18.0153 g/mol.

Finally, we subtract the mass of water vapor from the initial mass of water (1.5 g) to find the mass of liquid water.

Let's calculate it step by step:

Step 1: Calculate the number of moles of water vapor

From the ideal gas law equation PV = nRT, we can rearrange it as n = PV/(RT).

Since the pressure of the system is atmospheric pressure (~1 atm), the volume is 1.00 L, the R is 0.0821 L · atm/(mol · K), and the temperature is 298 K, we can calculate:

n = (1.00 atm × 1.00 L) / (0.0821 L · atm/(mol · K) × 298 K)

n ≈ 0.0401 mol

Therefore, there are approximately 0.0401 moles of water vapor in the vessel.

Step 2: Calculate the mass of water vapor

Using the molar mass of water (18.0153 g/mol), we can calculate the mass of the water vapor:

Mass of water vapor = Number of moles × Molar mass

Mass of water vapor = 0.0401 mol × 18.0153 g/mol

Mass of water vapor ≈ 0.722 g

Therefore, there are approximately 0.722 grams of water vapor in the vessel.

Step 3: Calculate the mass of liquid water

Finally, we can subtract the mass of water vapor from the initial mass of water to find the mass of liquid water:

Mass of liquid water = Initial mass of water - Mass of water vapor

Mass of liquid water = 1.5 g - 0.722 g

Mass of liquid water ≈ 0.778 g

Therefore, approximately 0.778 grams of liquid water would be present in the vessel.