Question

Enter your answer in the provided box. A 1.70−g sample of water is injected into an evacuated 4.50−L flask at 40.0° C. What percentage of water will be vapor when the system reaches equilibrium? Assume ideal behavior of water vapor and that the volume of liquid water is negligible. The vapor pressure of water at 40.0° C is 55.3 mmHg. An answer should be in percent.

Answer 1

Answer – We are given, mass of water = 1.70 g , volume = 4.5 L

T = 40.0 +273.15 = 313.15 K and at this temperature pressure P = 55.3 mm Hg

We know need to calculate the mass of water in the vapor phase.

Using the Ideal gas law, we can calculate moles of water in vapor phase.

PV = nRT

We need P in atm , so

760 mm Hg = 1 atm

55.3 mm Hg = ?

= 0.0728 atm

PV = nRT

n = P V / R T

= 0.0728 atm X 4.50 L / 0.0821 L.atm .mol^-1 .K ^-1 X 313.15 K

= 0.0127 moles of water

Mass of water in vapor phase = 0.0127 moles X 18.016 g/mol

                                               = 0.229 g

Percent of water vaporized = 0.229 g / 1.70 g X 100 %

                                          = 13.5 %

The 13.5 percentage of water will be vapor when the system reaches equilibrium