Question

A mixture of helium and xenon gases at a total pressure of 675 mm Hg contains helium at a partial pressure of 429 mm Hg. If the gas mixture contains 0.439 grams of helium, how many grams of xenon are present? g Xe

a mixture of helium and xenon gases at a total pressure of 675 mmHg contains helium at a partial pressure of 429 mmHg. if thr gas mixture contains 0.439 grams of helium how many grams of xenon are present

Answer 1

Solution :- Cliven that:: Total Pressure = 6 15 mmHg Partial Aress wrt of Helium = 429 mmHg Mass of Helium = 0.4399 moler mass of He = 49/mo so moles of she - 0.4399 - 0.1097.5 mol e 4g/mol we assume that's mole of xe = (hte) we know thati- Partiail. Pressure = Total Pressure (P) x mole fraction of gas. . Then mole fraction for He= mole of Helium [mole of Helium (he) + mole of xenon nxe) ] ° ~He = "H. Mhet nxe Put the value in above formula 429 mmHg = 675 mmHg x 0-10975 mol 0.10975 molt nexe) 4.29 mmHg 10.10975 mol 675 mm Hg (0.10975 molt nxe) 01635 = : 0.10975 mol 0:10975 + nxe 0.635(0.10975+%e) = 0.10975 mol => 6.06969+hxe0.635 0.10975 mol Mx 0.6"35= (0-10975-0.06969) nye 0635 = 0.04006 nxe = 0.04006 0.635 onxe = 0,06308 mulaio mass of xenon = 131.29 g/mole. So mass of Xe = 0.06308X131.29 g/mol 4ss / Xe - 8:379 3*