A gas mixture contains each of the following gases at the indicated partial pressures: N2, 315 torr; O2, 134 torr; and He, 219 torr?

What is the total pressure of the mixture? What mass of each gas is present in a 2.15 L sample of this mixture at 25.0 C?

Answer

total pressure = partial pressure due to N2 + partial pressure due to O2 + partial pressure due to He = 315 + 134 + 219 = 668 torr or 668/760 = 0.879 atm ( as 1 atm = 760 torr) now PV = nRT in order to calculate total mole of gas P = 0.879 atm V = 2.15 L n = total no.of moles of all three gases = ? R = 0.0821 L atm/K/mole T = 25 + 273 = 298 K 0.879 X 2.15 = n X 0.0821 X 298 1.889 = n X 24.466 n = 1.889/24.466 = 0.077 now partial pressure = mole fraction X total pressure partial pressure of N2 = mole fraction of N2 X total pressure 315 = no.of moles of N2/total no.of moles X 668 315 = no.of moles of N2/ 0.077 X 668 no.of moles of N2 = 315 X 0.077 / 668 = 0.036 molar mass of N2 = 28 g/mole so amount of N2 = 28 X 0036 = 1.008 g similalrly …partial pressure of O2 = mole fraction of O2 X total pressure 134 = no.of moles of O2/0.077 X 668 no.of moles of O2 = 134 X 0.077 / 668 = 0.015 molar mass of O2 = 32 g/mole so amount of O2 = 32 X 0.015 = 0.48 g and finally….partial pressure of He = mole fracion of He X total pressure 219 = no.of moles of He/0.077 X 668 no.of moles of He = 219 X 0.077/668 = 0.025 molar mass of He = 4 g/mole amount of He = 0.025 X 4 = 0.1 g

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