the vapor pressure of water at 25 C is 27.66 torr. is 0.5 g water is enclosed in a 1.5 L sealed container, will there be nay liquid. If yes, what mass?
I think you meant to type "If" instead of "is" at the start of your second sentence. I assume that the temperature of the container is 25 C.
27.66 torr = 27.66/760 = 0.036 atm
1 mole occupies 24.45 l at 25
The maximum amount of water that can be in gas phase at that temperature is
(number of moles of vapor)*(molar mass)
=(27.66/760)*(1.5/24.45)*18 g/mole
= 0.04 g
That leaves 4.96 g in liquid phase
To determine if there will be any liquid water in the sealed container, we need to compare the vapor pressure of water at 25°C to the actual vapor pressure inside the container.
Given:
Vapor pressure of water at 25°C = 27.66 torr
Mass of water = 0.5 g
Volume of the container = 1.5 L
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature
First, let's calculate the number of moles (n) of water:
We know the mass of water (0.5 g) and the molar mass of water (18 g/mol).
Number of moles (n) = Mass / Molar mass
= 0.5 g / 18 g/mol
= 0.02778 mol
Now, let's calculate the pressure (P) inside the container using the ideal gas law equation and rearranging it to solve for pressure:
P = nRT / V
Substituting the values we have:
P = (0.02778 mol) * (0.0821 L * atm / mol * K) * (298 K) / (1.5 L)
P ≈ 1.07 atm
To convert this pressure to torr, we use the conversion factor 1 atm = 760 torr:
P ≈ 1.07 atm * 760 torr / 1 atm
P ≈ 812.02 torr
Comparing the calculated pressure (812.02 torr) to the vapor pressure of water at 25°C (27.66 torr), we can conclude that the pressure inside the container is much higher than the vapor pressure of water. This means that all the water will exist as vapor, and there will be no liquid water in the container.
Therefore, the mass of liquid water in the container will be zero grams.