How many grams of water vapor (H2O) are in a 10.2 liter sample at 0.98 atmospheres and 26°C? Show all work used to solve this problem.
Look up the vapor pressure of H2O at 26 C and plug that into P in PV = nRT and solve for n = number of moles. Then n = grams/molar mass, solve for grams.
To solve this problem, we can use the ideal gas law equation: PV = nRT.
First, let's convert the given volume from liters to moles using the ideal gas law equation.
Step 1: Convert volume from liters to cubic meters
10.2 liters / 1000 = 0.0102 cubic meters
Step 2: Convert temperature from Celsius to Kelvin
26°C + 273.15 = 299.15 K
Step 3: Convert pressure from atmospheres to pascals
0.98 atmospheres * 101325 = 100130.5 pascals
Step 4: Substitute the given values into the ideal gas law equation
(100130.5 pascals)(0.0102 cubic meters) = n(8.314 J/mol·K)(299.15 K)
Solving for n, we get:
n = (100130.5 pascals * 0.0102 cubic meters) / (8.314 J/mol·K * 299.15 K)
n ≈ 0.0403 moles
Now that we know the number of moles (n) of water vapor, we can use the molar mass of water (H2O) to calculate the mass in grams.
The molar mass of water (H2O) is:
(2 hydrogen atoms * 1.008 g/mol) + (1 oxygen atom * 16.00 g/mol) = 18.02 g/mol
To find the mass of water vapor, we multiply the number of moles (n) by the molar mass:
0.0403 moles * 18.02 g/mol ≈ 0.726 g
Therefore, there are approximately 0.726 grams of water vapor in the 10.2 liter sample at 0.98 atmospheres and 26°C.