1. The problem statement, all variables and given/known data The pressure in a bulb dropped from 2000 to 1500 mm of mercury in 50 minutes when the contained oxygen leaked through a small hole. The bulb was then evacuated. A mixture of oxygen and another gas of molecular weight 72 in the molar ratio of 1 : 1 at a total pressure of 6000 mm of mercury was introduced. Find the molar ratio of the two gases remaining in the bulb after a period of 70 minutes. 2. Relevant equations 3. The attempt at a solution Since the evacuated bulb contains a mixture of oxygen and another gas in the molar ratio of 1 : 1 at a total pressure of 6000 mm, the partial pressure of each gas is 3000 mm. The drop in the pressure of oxygen after 70 minutes = 500/50*70 = 700 mm of Hg ∴ After 70 minutes, the pressure of oxygen = 3000-700=2300 mm of Hg Let the rate of diffusion of other gas be rn, then Rate of O2/Rate of gas =1.5 (graham's formula) ∴ Drop in pressure for the other gas = 700/1.5=1400/3 mm Hg ∴ pressure of the other gas after 70 minutes = 3000 – 1400/3 mm = 7600/3 mm of Hg Molar ratio = Moles of unknown gas/Moles of O2 = (7600/3)/(2300) = 76/69 [Partial pressure ∝ mole fraction] But the answer is 39/46. Where am I going wrong?