Question

A 500.0-ml flask contains 1.53 mg of oxygen and 0.23 mg of helium at 12°C. Calculate the partial pressures of oxygen and helium in the flask.

Answer 1

The partial pressure of oxygen in the flask is approximately 0.218 atm, and the partial pressure of helium is approximately 0.032 atm.

Step-by-step explanation:

To calculate the partial pressures of oxygen and helium, we first need to convert the mass of each gas to moles using the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

1. **Convert Oxygen Mass to Moles:**

- Molecular weight of oxygen
`(O_2\)`: 32 g/mol.

- Moles of oxygen
`(\(n_(O_2)\))` = Mass of oxygen / Molecular weight of oxygen.

-
`\(n_(O_2) = 0.00153 g / 32 g/mol = 4.78125 * 10^(-5)\) mol.`

2. Convert Helium Mass to Moles:

- Molecular weight of helium (He): 4 g/mol.

- Moles of helium
`(\(n_(He)\))` = Mass of helium / Molecular weight of helium.

-
`\(n_(He) = 0.00023 g / 4 g/mol = 5.75 * 10^(-5)\) mol.`

3. Calculate Partial Pressures:

- Total moles of gas
`(\(n_(total)\)) = \(n_(O_2) + n_(He) = 4.78125 * 10^(-5) + 5.75 * 10^(-5) = 0.0001033125\) mol.`

- Using
`\(PV = nRT\) and rearranging for pressure (\(P\)), \(P_(O_2) = (n_(O_2)/n_(total)) * P_(total)\) and \(P_(He) = (n_(He)/n_(total)) * P_(total)\).`

- Substituting values,
`\(P_(O_2) = (4.78125 * 10^(-5)/0.0001033125) * 1 \, \text{atm} \approx 0.218 \, \text{atm}\) and \(P_(He) = (5.75 * 10^(-5)/0.0001033125) * 1 \, \text{atm} \approx 0.032 \, \text{atm}\).`