Question

A mixture of argon and carbon dioxide gases at a total pressure of 866 mm Hg contains argon at a partial pressure of 507 mm Hg. If the gas mixture contains 7.53 grams of argon, how many grams of carbon dioxide are present?

Answer 1

Solution. Given that :- Tote Pressure of gases(P)= 366 mmHg Partial pressure of Argon Par) =507 MMHG mass of Argon (m) = 7.53g molar mass of Argon (m) = 39.95gimol So mote of Argon (nar) = 7.539 39.95g/mol nar-0.1885 mol Let the mole of co2 = ncos So To teel mole of gases = (nart hoz) so mole fraction of Argon (TA) = 1 MAY hartno, ri Partial Pressure = Total Pressure x mole fraction Partial pressure of Argon = 507 mmHg Put will above values in the equations 507mm Hg = 866 mm Hg 10.1885 mole 0:1885 moletno 507mm Hg 0.1885 mole 0.1885 mole+nco, 866 mm 0.5855 = 0.1885 mole 0.1885 mole +nco

0.5855(0-1885 molethcox) = 01885 mole 0.1104 +0.5855 nco, = 0.1885 mole 0.5855 No, = 6.1885-001104) mole 0.5855860, 0:0781 mole

nco2 = 0.1334 mole molar mass of 202=449/mole so mass of CO2 = 0.1334 molex 44 g ) 5,879 mole mass of CO2 = 5.87 g Ans.