Question

In a sealed 250 mL flask, 0.0403 g of liquid water is in equilibrium with its vapor at 70°C. Calculate the percentage of the total mass of water that is in the vapor phase.

Water data:       ρ = 1.00 g mL^–1; p*(70°C) = 0.308 atm

Answer 1

Volume of flask = 250 mL = 0.250 L

volume of vapor = volume of flask = 0.250 L

vapor pressure of water = 0.308 atm

temperature of system = 70 ^oC = 343 K

moles of water in vapor = (pressure * volume) / (R * temperature)

where R = gas constant = 0.0821 L-atm/mol-K

moles of water in vapor = [(0.308 atm) * (0.250 L)] / [(0.0821 L-atm/mol-K) * (343 K)

moles of water in vapor = 2.73 x 10^-3 mol

mass of water in vapor = (moles of water in vapor) * (molar mass of water)

mass of water in vapor = (2.73 x 10^-3 mol) * (18.0 g/mol)

mass of water in vapor = 0.0493 g

Total mass = mass of water in vapor + mass of water in liquid

Total mass = (0.0493 g) + (0.0403 g)

Total mass = 0.0896 g

percentage of water in vapor phase = (mass of water in vapor / total mass) * 100

percentage of water in vapor phase = (0.0493 g / 0.0896 g) * 100

percentage of water in vapor phase = 55.0 %