Answer:- partial pressure of Kr = 0.306 atm, partial pressure of oxygen = 0.264 atm and partial pressure of carbon dioxide = 0.396 atm
Total pressure is 0.966 atm
Solution:- moles of Kr = 21.7 g x (1mol/83.8g) = 0.259 mol
moles of oxygen = 7.18 g x (1mol/32g) = 0.224 mol
moles of carbon dioxide = 14.8 g x (1mol/44g) = 0.336 mol
Volume of container = 23.1 L and the temperature is 59 + 273 = 332 K
From ideal gas law equation, P = nRT/V
partial pressure of Kr = (0.259 x 0.0821 x 332).23.1 = 0.306 atm
partial pressure of oxygen = (0.224 x 0.0821 x 332)/23.1 = 0.264 atm
partial pressure of carbon dioxide = (0.336 x 0.0821 x 332)/23.1 = 0.396 atm
Total pressure of the gas mixture = 0.306 atm + 0.264 atm + 0.396 atm = 0.966 atm
