Final answer:
The sample of chlorine gas (Cl2) contains more molecules than the sample of Argon (Ar) because it has a greater number of moles when comparing both samples through calculations based on the Ideal Gas Law for Argon and molar mass for chlorine.
Step-by-step explanation:
To determine which sample contains more molecules, we need to calculate the number of moles for each given substance and then use Avogadro's number to compare the number of molecules. For the sample of Argon (Ar), we use the Ideal Gas Law: PV = nRT.
To find n (number of moles of Ar), we rearrange the ideal gas law to n = PV/RT. We need the pressure (P) in atmospheres, which is 1111 mm Hg / 760 mm Hg per atm = 1.461 atm. We have V as 2.84 L and R, the ideal gas constant, as 0.0821 L·atm/(mol·K). The temperature (T) must be in Kelvin, so we convert 85.0°C to 358.15 K.
n = (1.461 atm)(2.84 L) / (0.0821 L·atm/mol·K)(358.15 K) = 0.099 mol of Ar
To determine the number of moles of chlorine gas (Cl2), we use its molar mass, which is about 70.9 g/mol. Moles of Cl2 = 10.78 g / 70.9 g/mol = 0.152 mol of Cl2.
Since one mole of any substance contains the same number of molecules (Avogadro's number, 6.022 x 10^23 molecules/mol), we can say that the sample with more moles will have more molecules. Therefore, the sample of Cl2 contains more molecules than the sample of Argon since it has more moles.