Question

The Henry's law constant (kh) for O2 in water at 20°C is 1.28 x 10-3 mol/(L'atm). (a) How many grams of O2 will dissolve in 4.75 L of H20 that is in contact with pure 02 at 1.00 atm? g 02 (b) How many grams of O2 will dissolve in 4.75 L of H20 that is in contact with air where the partial pressure of O2 is 0.209 atm? g 02

Answer 1

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-3 kn = 1.28 x 10 mal 2x atm solublity (inmale) = Volume of x kHX Solvent Pressure af Oz -3 = 4,75 % x 1.28x10 mal x loo atm ť atm males of O2 = 0.00608 So Mass a moles X M.W 0.00608 male x 32 GM = mall Too 1945. gm Ang b) -3 solublity in male - 4.751x10 28 x 10 oodal x 0.209 atm Lotm = 0.00127 male 3 Mass of Oz z 0.00127 male x 3 2 gm male = 0.04064 gm Ans