
Respuesta :
Answer:
Mole fraction of Nâ‚‚O Â = 0.330
Mole fraction of SFâ‚„ = 0.669 Â
Pressure (Nâ‚‚O) = Â 39.12 kPa
Pressure (SFâ‚„) = 79.12 Pa
Total Pressure = 118.25 kPa Â
Explanation:
Given Â
Given mass of dinitrogen difluoride Nâ‚‚0 =5.53 g
Given mass of sulphur hexafluoride =17.3 g
volume of tank V= 8 L= 0.008 ml
R ideal gas constant =  8.31  J / mol·K
Temperature = 26.9 C = 299.9 k
Pressure = ?
Number of moles = ? Â
Number of moles of Nâ‚‚O in tank = given mass /molar mass =5.53 / (14.0 x 2 + 16.0) = 0.1256 mol
Number of moles SFâ‚„ in tank = given mass /molar mass 17.3 / (32.1 + 19.0 x 4) = 0.254 mol
Mole fraction of Nâ‚‚O = Number of moles of Nâ‚‚0/total number of moles Â
                     = 0.1256 / (0.1256+0.254) = 0.330
Mole fraction of SFâ‚„ = 0.254 / (0.1256+0.254= 0.669 Â
For Pressure Â
PV=nRT
Pressure (Nâ‚‚O) = (0.1256)(8.31)(299.9) / (0.008) = 187.22 = Â 39127.05 Pa = 39.12 kPa
Pressure (SFâ‚„) = (0.254)(8.31)(299.9) / (0.008) = 79126.36 Pa= 79.12 Pa
Total Pressure = 39.12 + 79.12 = 118.25 kPa Â