a 3m x 3m x 4m room contains air with a moisture content of 11.6 gram H20/dry air at 1 atm (absolute) pressure and 25 degree Celsius. the dry air component can be assumed to have a composition of 21 mol% oxygen and 79 mol% nitrogen. 1) calculate the mass fraction of water vapour in the room.2 2) mole fraction of water vapour in the room?

Question

cuanchi · Answer

moisture content of 11.6 g h20/dry air =moisture content of 11.6 g H2O % dry air ???

anonymous · Answer

any one there to solve above question?

cuanchi · Answer

please tell me what do you mean with this, I can not help you without that!

"11.6 g h20/dry air"

anonymous · Answer

it means 11.6 gram H2O/kg dry air

cuanchi · Answer

molar mass of dry air
(0.21*32+0.79*28)= 28.84g/mol

11.6 g H2O /18g = 0.64 mol H2O/kg dry air

1000g dry air/28.84g =34.67 mol dry air

11.6 g H2O/kg of dry air = 0.64 mol H2O/34.67 mole Dry air= 
1mol H2O/54 mol dry air = 54mol dry air/mol of H2O
0.0185 mol H2O/mol of dry air

you have 1mol of H2O every 54 mole of dry air or every 55 mole of gas total volume of your room

the total volume of your room is 3*3*4= 36 cubic meters= 36000L
the molar volume of a ideal gas in STP conditions is 22.4L
then you can calculate how many moles of gas you have in the room
36000L/22.4L= 1607.14 moles of gas 
for every 55 moles of gas one is H2O
1607.14/55= 29.22 moles of H2O in the room => 29.22*18= 252.97 g H2O


    29.2 moles of H2O in the room     = 29.22*18           = 252.97 g H2O
1576.8 moles of dry air in the room  = 1576.8*28.84= 45474.9 g dry air
1607 moles of gas in the room                                   =45727.882 g

45727.882/1607 =28.46g/mol gas in the room