Answer :

Given


Temperature T=300K


Relative humidity=20%


Volume of room= 50m3


Saturation vapor pressure at 300 K is 3.3 kPa=3.3Pa.




P=0.23.3103=660Pa


Molar mass of water H2O =2+16=18g


Applying ideal gas equation


PV=nRT


Where V= volume of gas


R=gas constant =8.3JK-1mol-1


T=temperature


n=number of moles of gas


P=pressure of gas.


Number of moles n=





m=238.55g


Mass of water vapor present in the room=238.55g.


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