The mass percent of Ar in the mixture is 93.53%. Below is why:
The molar mass of Ar (Argon) is 39.948 g/mol and the molar mass of CH4 (Methane) is 16.0425 g/mol.
The mole fraction of Ar is given as 0.853. This means that 85.3% of the total moles of gas in the mixture are Ar, and the remaining 14.7% are CH4.
The total moles of gas in the flask can be calculated using the ideal gas law equation, PV = nRT, where:
- P is the pressure,
- V is the volume,
- n is the number of moles,
- R is the ideal gas constant, and
- T is the temperature.
Rearranging the equation to solve for n gives n = PV/RT.
Substituting the given values into the equation gives:
Total moles of gas = (1.53 atm * 9.08 L) / (8.206 x 10^-2 L.atm.K-1.mol-1 * 300 K) = 0.853 * total moles of Ar + (1 - 0.853) * total moles of CH4.
Solving the above equation gives:
Total moles of Ar = 0.853 * total moles of gas = 20.999154 moles,
Total moles of CH4 = (1 - 0.853) * total moles of gas = 3.618846 moles.
The mass of each gas can then be calculated by multiplying the number of moles by the molar mass:
Mass of Ar = total moles of Ar * molar mass of Ar = 20.999154 moles * 39.948 g/mol = 839.161 g,
Mass of CH4 = total moles of CH4 * molar mass of CH4 = 3.618846 moles * 16.0425 g/mol = 58.080 g.
Finally, the mass percent of Ar:
Mass percent of Ar = (mass of Ar / (mass of Ar + mass of CH4)) * 100 = (839.161 g / (839.161 g + 58.080 g)) * 100 = 93.5273247%.
In other words, the mass percent of Ar in the mixture is approximately 93.53%.