Answer
See the answer below.
Work Step by Step
From appendix G, at 23$°C$, the vapor pressure of water is: 21.1 mmHg.
Partial pressure: $55/100×21.1=11.61\ mmHg$
Assuming that the room pressure is 1 atm:
Mol fraction: $11.61/760×100\%=1.52\%$
Mass per mol of air: $1.52/100×18.015=0.274\ g/mol$
Molar volume of air: $V/n=0.082×296\ K/1\ atm=24.29\ L/mol$
Mass of water per liter of air:
$0.274\ g/mol\div 24.29\ L/mol=0.0113\ g/L$
Volume of the room:
$4.5\ m^2×3.5\ m=15.75\ m^3=15750\ L$
Mass of water in the room:
$0.0113\ g/L×15750\ L=178\ g$
