Calculate the mass of water vapour present in a room of volume 400 m3 that contains air at 27 °c on a day when the relative humidity is 60 per cent. hint: relative humidity is the prevailing partial pressure of water vapour expressed as a percentage of the vapour pressure of water vapour at the same temperature (in this case, 35.6 mbar).

Mass of water = 6251. 706g or 6.25Kg

Explanation:

Relative humidity = (actual vapor pressure/saturation vapor pressure) * 100%

Actual vapor pressure, Pw = relative humidity * saturation vapor pressure

Pw = 60% * (35.6 *0.001)atm = 0.0216atm

Note: 1mbar = 0.001atm

Using the ideal gas equation: PV=nRT; where P = Pw= 0.02136atm, V= (400

* 1000)dm^3, R= 0.082 atmdm^3/kmol, T= (27+273)K, n = number of moles

Note: 1m^3 = 1000dm^3,R is the molar gas constant.

Making n subject of the formula, n = PV/RT

n= (0.02136 * 400000)/(0.082 * 300) = 347.317 moles

Mass (g) = number of moles (n) * molar mass

molar mass of water=18g

Mass of water = 347.317 * 18 = 6251. 706g or 6.25Kg

the formula would be mgso4

mgso[subscript]4

your answer would be a. 12.5 kg