Question

Please answer with clear steps and explanation, thankyou

Practice Exercise 1 In a system for separating gases, a tank containing a mixture of hydrogen and carbon dioxide is connected to a much larger tank where the pressure is kept very low. The two tanks are separated by a porous membrane through which the molecules must effuse. If the initial partial pressure of each gas is 5.00 atm, what will be the mole fraction of hydrogen in the tank after the partial pressure of carbon dioxide has declined to 4.50 atm? (a) 52.1% (b) 37.2% (c) 32.1% (d) 4.68% (e) 27.4%

Answer 1

The rate of effusion is inversely proportional to the square root of mass

Molar mass of CO2 = Molar mass of C + 2 * Molar mass of O = 12 + 2 * 16 = 44 g/mol

44 1/2 Mco, y1/2 = ( )2 4.69 TH2 2 Мн, rco

гн, — 4.69тсо,

Change in pressure of CO2 = 5.0 atm - 4.5 atm = 0.5 atm

4.69 Change Change in pressure of H2 in pressure of CO2 77 E 4.69 0.5 2.34

Final Pressure of H2 = 5.0 - 2.345 = 2.655 atm

mole-fraction of hydrogen = (pressure exerted by hydrogen)/(pressure exerted by hydrogen + pressure exerted by CO2) * 100 = 2.655/(2.655+4.50) * 100 = 37.2%

Correct answer is option (b)

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