A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will be vapor when equilibrium is reached? (the vapor pressure of water at 65°C is 187.5 mmHg)

50.4 % of the water will be vapor

Explanation:

Step 1: Data given

Mass of water = 1.68 grams

volume of the flask = 5.3 L

Temperature = 65°C

Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

Step 2: Calculate moles of H2O

p*V=n*R*T

⇒ p = the pressure of water = 0.2467 atm

⇒ V = the volume of the flask = 5.3 L

⇒ n = moles of water

⇒ R = gas constant = 0.08206 L*atm / K*mol

⇒ T = the temperature = 65°C = 338 Kelvin

n = (p*V) / (R*T)

n = (0.2467 * 5.3) / (0.08206 * 338)

n = 0.047 moles

Step 3: Calculate mass of water

Mass of water = moles of water * molar mass of water

Mass of water = 0.047 moles * 18.02 g/mol

Mass of water = 0.84694 grams

Step 4: Calculate the percent of water vaporized

% = (0.84694 grams/1.68 grams) * 100%

% = 50.4%

50.4 % of the water will be vapor