11. It took 1 min. and 37 sec for a given volume of chlorine to effuse through a
ID: 591521 • Letter: 1
Question
11. It took 1 min. and 37 sec for a given volume of chlorine to effuse through a pinhole under given conditions of temperature and pressure. How long will it take for the same volume of water vapor to effuse through the same hole under the same conditions? 12. [Bonus, More Challenging!] A mixture of methane, CH4, and acetylene, C2H2, occupied a certain volume at a total pressure of 63 mm Hg. The sample was burned to CO2 and H20, and the CO2 alone was collected and its pressure was found to be 96 mm Hg in the same volume and at the same temperature as the original mixture. What fraction of the gas was methane?Explanation / Answer
11) Let we have V mL of both chlorine and water vapor. The rate of diffusion of chlorine is given as
(V mL)/(1 min 37 s) = (V mL)/(97 s) [since 1 min = 60 s].
Let t s be the time taken for V mL of water vapor to diffuse through the same pinhole; then the rate of diffusion of water vapor is (V mL)/(t s).
As per Graham’s law of diffusion,
Rate of diffusion of chlorine/Rate of diffusion of water vapor = (molar mass of water vapor)/(molar mass of chlorine)
Molar mass of water vapor = (2*1.008 + 1*15.9994) g/mol = 18.0154 g/mol; molar mass of chlorine = (2*35.453) g/mol = 70.906 g/mol.
As per the problem,
(V mL/97 s)/(V mL/t s) = (18.0154 g/mol)/(70.906 g/mol)
====> t/97 = 0.25407
====> t/97 = 0.50406
====> t = 0.50406*97 = 48.89382 49
The rate of diffusion of water vapor through the same pinhole is 49 s (ans).
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