If 10.00 g of water are introduced into an evacuated flask of volume 2.500 L at 65 degreed celcius, calculate the mass of water vaporized. (Hint: Assume that the volume of the remaining liquid water is negligible; the vapor pressure of water at 65 degrees celsius is 187.5 mmHg.)

To calculate the mass of water vaporized, we need to determine the amount of water that has evaporated from the flask. We can do this by using the ideal gas law, which relates the pressure, volume, temperature, and amount of gas. Here's how we can solve this problem step by step:

1. Convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 65 + 273.15 = 338.15K

2. Convert the vapor pressure from mmHg to atm:
1 atm = 760 mmHg
P(atm) = P(mmHg) / 760
P(atm) = 187.5 / 760 = 0.24605 atm

3. Apply the ideal gas law equation:
PV = nRT

P: Pressure in atmospheres (0.24605 atm)
V: Volume in liters (2.500 L)
n: Amount of substance in moles (unknown)
R: Ideal gas constant (0.08206 L·atm/(mol·K))
T: Temperature in Kelvin (338.15 K)

4. Rearrange the equation to solve for the amount of substance (moles):
n = PV / RT

n = (0.24605 atm * 2.500 L) / (0.08206 L·atm/(mol·K) * 338.15 K)
n = 0.024003 mol

5. Convert moles of water to grams:
The molar mass of water (H₂O) is approximately 18.015 g/mol.

Mass = n * molar mass
Mass = 0.024003 mol * 18.015 g/mol
Mass ≈ 0.432 g

Therefore, the mass of water vaporized is approximately 0.432 grams.

To calculate the mass of water vaporized, we can use the ideal gas law and the vapor pressure of water at 65 degrees Celsius.

Step 1: Convert the given vapor pressure of water from mmHg to atm.
Given: Vapor pressure of water at 65 degrees Celsius = 187.5 mmHg
Note: 1 atm = 760 mmHg

To convert mmHg to atm, we divide the value by 760.
187.5 mmHg / 760 mmHg = 0.2467 atm

Step 2: Calculate the number of moles of water vapor using the ideal gas law. The ideal gas law equation is:
PV = nRT

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

Given:
P = 0.2467 atm (vapor pressure of water at 65 degrees Celsius)
V = 2.500 L (volume of the flask)
T = 65 degrees Celsius = 65 + 273 = 338 Kelvin

Using the ideal gas law:
0.2467 atm * 2.500 L = n * 0.0821 L.atm/mol.K * 338 K

Solving for n:
n = (0.2467 atm * 2.500 L) / (0.0821 L.atm/mol.K * 338 K)
n ≈ 0.0278 mol

Step 3: Calculate the mass of water vaporized using the molar mass of water (H2O).
The molar mass of H2O = 18.015 g/mol.

Using the number of moles calculated in step 2:
mass = n * molar mass
mass = 0.0278 mol * 18.015 g/mol
mass ≈ 0.500 g

Therefore, the mass of water vaporized is approximately 0.500 grams.