How many grams of O2 will dissolve in 2.50 L of water that is in contact with O2 at a partial pressure of 0.209 atm and 20 °C? The Henry’s Law constant for O2 (kH) in water at 20 °C is 1.28 × 10−3 mol L−1 atm−1.
We can use Henry's Law to solve for the amount of O2 that will dissolve in water:
O2(g) <=> O2(aq)
where O2(g) is the gas phase and O2(aq) is the dissolved phase.
According to Henry's Law, the amount of O2 that will dissolve in water is directly proportional to its partial pressure:
O2(aq) = kH * P(O2)
where kH is the Henry's Law constant and P(O2) is the partial pressure of O2.
Plugging in the given values, we get:
O2(aq) = (1.28 x 10^-3 mol L^-1 atm^-1) * (0.209 atm)
O2(aq) = 2.68 x 10^-4 mol L^-1
To convert this to grams, we need to use the molar mass of O2:
molar mass of O2 = 32 g/mol
So, the mass of O2 that will dissolve in 2.50 L of water is:
mass of O2 = (2.68 x 10^-4 mol L^-1) * (32 g/mol) * (2.50 L)
mass of O2 = 0.0214 g