A mixture of krypton and oxygen gases, in a 6.92 L flask at 34 °C, contains 26.4 grams of krypton and 6.16 grams of oxygen. The partial pressure of oxygen in the flask is

atm and the total pressure in the flask is
atm.

1 answer

To find the partial pressure of oxygen, we first need to calculate the moles of each gas present in the mixture.

1. Calculate the moles of krypton:

molar mass of krypton (Kr) = 83.80 g/mol
moles of krypton = mass of krypton / molar mass of krypton
moles of krypton = 26.4 g / 83.80 g/mol
moles of krypton = 0.315 mol

2. Calculate the moles of oxygen:

molar mass of oxygen (O2) = 32.00 g/mol
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 6.16 g / 32.00 g/mol
moles of oxygen = 0.193 mol

3. Calculate the total moles of gas in the mixture:

total moles = moles of krypton + moles of oxygen
total moles = 0.315 mol + 0.193 mol
total moles = 0.508 mol

4. Now, we can calculate the mole fraction of oxygen:

mole fraction of oxygen = moles of oxygen / total moles
mole fraction of oxygen = 0.193 mol / 0.508 mol
mole fraction of oxygen = 0.380

5. Finally, we can find the partial pressure of oxygen using the mole fraction and total pressure:

partial pressure of oxygen = mole fraction of oxygen * total pressure
partial pressure of oxygen = 0.380 * X atm

Given that total pressure is X atm, we need to find the value of X.

Using the ideal gas law:
PV = nRT
X atm * 6.92 L = 0.508 mol * 0.0821 atm*L/mol*K * 34 + 0.508 mol * 0.0821 atm*L/mol*K * 273

X * 6.92 = 0.508 * 0.0821 * (34 + 273)
X * 6.92 = 19.84

X = 19.84 / 6.92
X = 2.87 atm

Therefore, the partial pressure of oxygen in the flask is:
partial pressure of oxygen = 0.380 * 2.87 atm
partial pressure of oxygen = 1.09 atm

The total pressure in the flask is 2.87 atm.