Question

A 19.3-g mixture of oxygen and argon is found to occupy a volume of 16.2 l when measured at 675.9 mmhg and 43.4oc. what is the partial pressure of oxygen in this mixture?

Answer 1

438.0 mmHg is the partial pressure of oxygen The ideal gas law is PV = nRT where P = pressure V = volume n = number of moles R = ideal gas constant (8.3144598 (L*kPa)/(K* mol) ) T = absolute temperature Converting the temperature from C to K gives 43.4 + 273.15 = 316.55 K Converting pressure from mmHg to kPa gives 675.9 mmHg * 0.133322387415 = 90.11260165 kPa Solving for n in the equation for the ideal gas law, gives PV = nRT n = PV/(RT) Substituting known values into the equation. n = PV/(RT) n = 90.11260165 kPa 16.2 L/(8.3144598 (L*kPa)/(K* mol) 316.55 K) n = 1459.824147 L*kPa / 2631.94225 (L*kPa)/(mol) n = 0.554656603 mol So we have 0.554656603 moles of gas particles. Now to determine how much of that is oxygen. Atomic weight oxygen = 15.999 g/mol Atomic weight argon = 39.948 g/mol Molar mass O2 = 2 * 15.999 = 31.998 g/mol Now we need to figure out how many moles of O2 gas and Ar will both add up to the number of moles of gas particles and have the proper mass. So M = number of moles of O2 0.554656603 - M = number of moles of Ar and M * 31.998 + (0.554656603 - M) * 39.948 = 19.3 Solve for M M * 31.998 + (0.554656603 - M) * 39.948 = 19.3 M * 31.998 + 22.15742198 - M * 39.948 = 19.3 -M * 7.95 + 22.15742198 = 19.3 -M * 7.95 = -2.857421977 M = 0.359424148 So we now know that we have 0.359424148 moles of oxygen gas out of a total of 0.554656603 moles of gas particles. So oxygen gas is providing: 0.359424148 / 0.554656603 = 0.648012024 = 64.8012024% of the total pressure of 675.9 mmHg. So the partial pressure is 675.9 * 0.648012024 = 437.9913271 mmHg = 438.0 mmHg