A glass of pure water is put outside for one hour. The air is polluted with 5g/m3 of SO3. The SO3 dissolves in the water and reacts with water to form H2SO4, keeping the SO3 concentration in the water at essentially zero. The reaction is slow enough that it does not enhance mass transfer other than keeping the bulk SO3 concentration in the water equal to zero Calculate: - the mass flux of SO3 into the water - the concentration of H2SO4 in the water after one hour. Assume the following: - Area of the water surface: 20cm2 - Volume of water in the glass: 0.2 liter - Molar mass SO3:80g/mol - Molar mass H2SO4:98g/mol - H=106 (Henry constant) - kL=2106m/s - kG=2103m/s