3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Dry air at a total pressure of 1.00 bar contains N2, O2 and Ar with partial pressures 0.78 bar, 0.21 bar and 0.0093 bar. At the temperature of the experiment the Henry's law constants (in the form p=KHx, x = mole fraction of the dissolved gas) in water are KH(N2) = 86 kbar, KH(O2) = 44 kbar and KH(Ar) = 40 kbar. Water is equilibrated with this air, and all the dissolved gas is then expelled by boiling and collected. The percentage (by moles) of O2 in the expelled gas is closest to ______. Exactly one option is correct.
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Answer: C
By Henry's law the mole fraction of each dissolved gas is xi=pi/KH,i (all xi refer to the same amount of water, so they are proportional to the amounts dissolved).
N2: 0.78/86000=9.07×10−6; O2: 0.21/44000=4.77×10−6; Ar: 0.0093/40000=2.3×10−7.
Percentage of O2 in the expelled gas =9.07+4.77+0.234.77×100=33.9%≈34%, option C.
Dissolved air is enriched in O2 (34% against 21% in air) because O2 has the smaller KH, that is, the higher solubility. Option A (12%) comes from multiplying instead of dividing by KH; option B ignores the different solubilities.
Difficulty: HardQuestion 2Level 3 · AdvancedNumerical answer
A bottle contains 500 mL of water (density 1.00 g mL−1) saturated with CO2 at 298 K under a CO2 partial pressure of 2.50 bar. The bottle is opened and the liquid equilibrates with air, in which the partial pressure of CO2 is 4.0×10−4 bar, at 298 K. Henry's law constant of CO2 in water at 298 K is KH = 1.67 kbar (in p=KHxCO2). Ignore any reaction of CO2 with water and any change in the amount of water. Molar masses: H2O = 18.0 g mol−1, CO2 = 44.0 g mol−1. What mass of CO2 leaves the liquid, in g? Give the answer correct to two decimal places.
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Answer: 1.83
Amount of water: nw=500/18.0=27.78 mol.
Before opening: x1=2.50/1670=1.497×10−3, so nCO2=1−x1x1nw=4.164×10−2 mol.
After opening: x2=4.0×10−4/1670=2.4×10−7, so n=6.7×10−6 mol (negligible).
CO2 released =(4.164×10−2−0.0007×10−2)×44.0=1.832 g ≈ 1.83 g.
(Using the dilute approximation n≈xnw gives 1.830 g, which also rounds to 1.83 g.)
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
Liquids A and B form an ideal solution. At temperature T, a liquid with mole fraction xA = 0.40 is in equilibrium with a vapour in which the mole fraction of A is yA = 0.60, the total vapour pressure being 300 mm Hg. At the same temperature, the mole fraction of A in the liquid that is in equilibrium with a vapour containing equal numbers of moles of A and B is ______. Exactly one option is correct.
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Answer: A
Raoult's law with Dalton's law: yAP=xApA∘ and yBP=xBpB∘.
pA∘=0.60×300/0.40=450 mm Hg; pB∘=0.40×300/0.60=200 mm Hg. (Check: 0.40×450+0.60×200=300.)
For a given vapour composition: P1=pA∘yA+pB∘yB=4500.5+2000.5, so P=276.9 mm Hg.
Then xA=yAP/pA∘=0.5×276.9/450=0.308=4/13≈0.31, option A.
The liquid is poorer in the more volatile A than the vapour, as it must be. Option C wrongly equates liquid and vapour compositions; D interchanges A and B.
Difficulty: Very hard