In: Chemistry
What is the mass of oxygen dissolved in a 7.0L fish
tank when pressure of air is 0.98atm and the mole fraction of
oxygen in air is 0.22. T=25 C degrees, Henry's constant KH for O2
is 1.3 x 10^-3 M/atm at 25 C degrees.
A) 62.9 mg
B) 123.6 mg
C) 48.9 mg
D) 285.4 mg
partial pressure of O2 = total pressure x mole fraction
= 0.98 x 0.22
= 0.2156 atm
molarity of O2 = KH x partial pressure
= 1.3 x 10^-3 x 0.2156
= 2.8028 x 10^-4 M
moles = molarity x volume
= 2.8028 x 10^-4 x 7.0
= 1.96 x 10^-3
mass of O2 = 1.96 x 10^-3 x molar mass
= 1.96 x 10^-3 x 32
= 0.0629 g
= 62.9 mg
answer : A) 62.9 mg