A 1.00 -L gas sample at \(100 .^{\circ} \mathrm{C}\) and 600 . torr contains 50.0\(\%\) helium and 50.0\(\%\) xenon by mass. What are the partial pressures of the individual gases?

First, we need to calculate the mass of each gas for the given mass percentages. We can assume a total mass of 100g for the mixture, which will make the calculation easier. Thus, 50% helium and 50% xenon would mean that there are 50g of helium and 50g of xenon in the mixture.

Next, we'll calculate the moles of each gas using their molar masses. The molar mass of helium is approximately 4g/mol, and the molar mass of xenon is approximately 131g/mol. For helium: Moles = mass / molar mass Moles of helium = \( \frac{50 g}{4 g/mol} \) = 12.5 mol For xenon: Moles of xenon = \( \frac{50 g}{131 g/mol} \) = 0.381 mol

Now, we'll use the ideal gas law to compute the partial pressures for each gas. The ideal gas law is given by: \( PV = nRT \) Where: P = pressure V = volume n = moles R = ideal gas constant (0.0821 L.atm/(mol.K) ) T = temperature in Kelvin First, we need to convert the given temperature to Kelvin: Temperature in Kelvin = 100°C + 273.15 = 373.15 K Also, we need to convert the total pressure from torr to atm: Pressure in atm = \( \frac{600 torr}{760 torr/atm} \) = 0.789 atm For helium: \( P_{He}V = n_{He}RT \) \( P_{He}(1.00 L)= (12.5 mol)(0.0821 L.atm/(mol.K))(373.15 K) \) \( P_{He} = \frac{(12.5 mol)(0.0821 L.atm/(mol.K))(373.15 K)}{1.00 L} \) \( P_{He} \approx 0.383 atm \) For xenon: \( P_{Xe}V = n_{Xe}RT \) \( P_{Xe}(1.00 L)= (0.381 mol)(0.0821 L.atm/(mol.K))(373.15 K) \) \( P_{Xe} = \frac{(0.381 mol)(0.0821 L.atm/(mol.K))(373.15 K)}{1.00 L} \) \( P_{Xe} \approx 0.406 atm \)

Now, we have the partial pressures for each gas. Partial Pressure of Helium: \( P_{He} \approx 0.383 atm \) Partial Pressure of Xenon: \( P_{Xe} \approx 0.406 atm \) These values are the partial pressures for the individual gases within the 1.00 L mixture at 100°C and 600 torr total pressure.